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Write the Lewis dot structures for:\
(a) $\mathrm{Cl}_{2}$\
(b) $\mathrm{H}_{2}\mathrm{O}$\
(c) $\mathrm{CCl}_{4}$. | First, determine the total number of valence electrons for each molecule. For $$\mathrm{Cl}_{2}$$, it is $$2 \times 7 = 14$$. For $$\mathrm{H}_{2}\mathrm{O}$$, it is $$(2 \times 1) + 6 = 8$$. For $$\mathrm{CCl}_{4}$$, it is $$4 + (4 \times 7) = 32$$. Next, identify the central atom and draw single bonds to the other at... | Chemistry 11th |
What is formal charge? How is it calculated? Explain with the help of ozone molecule. | First, draw the Lewis structure for ozone ($$\text{O}_3$$). It consists of a central oxygen atom, which is double-bonded to one oxygen atom and single-bonded to another. The overall molecule is neutral. The formula for formal charge is: $$FC = V - N - \frac{B}{2}$$, where $$V$$ is the number of valence electrons, $$N$$... | Chemistry 11th |
What are odd-electron molecules? Why do they not follow the octet rule? | The octet rule states that atoms are most stable when they have eight electrons in their valence shell. Odd-electron molecules are molecules that have an odd total number of valence electrons. They are also known as free radicals. An octet is an even number (8). It is impossible to distribute an odd number of total ele... | Chemistry 11th |
What is expanded octet? Which elements can show expanded octet and why? | The octet rule states that atoms tend to bond in such a way that they each have eight electrons in their valence shell, similar to a noble gas. An expanded octet is an exception to this rule, where a central atom in a molecule has more than eight valence electrons. Elements that can exhibit an expanded octet are those ... | Chemistry 11th |
What is lattice enthalpy? How does it affect the stability of ionic compounds? | Lattice enthalpy ($$\Delta H_{LE}$$) is defined as the energy required to completely separate one mole of a solid ionic compound into its constituent gaseous ions. The process is endothermic, meaning energy must be supplied, so lattice enthalpy values are always positive. For example: $$\mathrm{NaCl(s)} \rightarrow \ma... | Chemistry 11th |
Define bond length. How is it related to covalent radii? | Bond length is the average, equilibrium distance between the centers of the nuclei of two atoms that are covalently bonded together. Covalent radius is defined as one-half the distance between the nuclei of two identical atoms joined by a single covalent bond. The bond length ($$d$$) between two different atoms, A and ... | Chemistry 11th |
Distinguish between covalent radius and van der Waals radius. | Covalent radius is half the distance between the nuclei of two atoms in a covalent bond. Van der Waals radius is half the distance between the nuclei of two non-bonded atoms in adjacent molecules. In a covalent bond, the electron clouds of the atoms overlap, pulling the nuclei closer together. For non-bonded atoms, the... | Chemistry 11th |
Define bond order. How is it related to bond strength and bond length? | Bond order is the number of chemical bonds between two atoms. It is calculated as half the difference between the number of bonding electrons and antibonding electrons. As bond order increases, the bond strength increases. This is because more electrons are holding the atoms together, requiring more energy to break the... | Chemistry 11th |
What is resonance? Explain with the help of ozone molecule. | Resonance is a way of describing delocalized electrons within certain molecules where the bonding cannot be expressed by a single Lewis structure. The ozone molecule ($$\text{O}_3$$) has 18 valence electrons. We can draw two possible Lewis structures. In one, a double bond exists between the central oxygen and the left... | Chemistry 11th |
Write the resonance structures of carbonate ion and explain why all C-O bonds are equivalent. | The carbonate ion, $$\text{CO}_3^{2-}$$, has a central carbon atom bonded to three oxygen atoms. The total number of valence electrons is 24. To satisfy the octet rule for all atoms, one C-O bond is a double bond, and two are single bonds. This arrangement can be drawn in three different ways. These three structures ar... | Chemistry 11th |
What are the important features of resonance structures? | Resonance describes a molecule where a single Lewis structure is insufficient to represent its true electron distribution. It involves delocalized electrons spread over multiple atoms. Resonance structures differ only in the placement of pi ($$\pi$$) electrons and lone pairs, not the position of atoms. The overall mole... | Chemistry 11th |
Clear the misconceptions associated with resonance. | Resonance is a method used to represent delocalized electrons when a single Lewis structure is inadequate. These multiple structures are called resonance structures or contributors. Misconception 1: Molecules do not rapidly switch or 'resonate' between different resonance structures. The actual molecule is a single, un... | Chemistry 11th |
Define dipole moment. What is its unit? How is it represented? | A dipole moment is the product of the magnitude of the charge and the distance of separation between the charges. It is a measure of the overall polarity of a molecule. The formula is $$\mu = q \times d$$, where $$\mu$$ is the dipole moment, $$q$$ is the magnitude of the separated charge, and $$d$$ is the distance betw... | Chemistry 11th |
Explain why $\mathrm{BF}_{3}$ has zero dipole moment despite having polar B-F bonds. | The electronegativity of Fluorine (3.98) is significantly higher than that of Boron (2.04). This difference causes each B-F bond to be polar, creating a partial negative charge on the fluorine atom and a partial positive charge on the boron atom. VSEPR theory predicts the shape of $$\mathrm{BF}_{3}$$. The central Boron... | Chemistry 11th |
State Fajans' rules for the covalent character of ionic bonds. | A small cation has a high charge density and can more effectively pull the electron cloud of the anion towards itself, increasing covalent character. A large anion is more easily polarized because its outermost electrons are further from the nucleus and less tightly held. This increased distortion enhances covalent cha... | Chemistry 11th |
Explain the repulsive interactions between electron pairs in order of decreasing repulsion. | The types of electron pairs are lone pairs (lp) and bonding pairs (bp). A lone pair is held by only one nucleus, so its electron cloud is spread out and occupies more space, causing greater repulsion. A bonding pair is shared between two nuclei, which confines its electron cloud to the region between the atoms, resulti... | Chemistry 11th |
Using VSEPR theory, predict the shape of $\mathrm{BeCl}_{2}$, $\mathrm{BCl}_{3}$ and $\mathrm{CH}_{4}$. | For $$BeCl_2$$, the central Be atom has 2 valence electrons, forming two single bonds with two Cl atoms. There are 2 electron domains and no lone pairs. The resulting shape is linear with a 180 bond angle. For $$BCl_3$$, the central B atom has 3 valence electrons, forming three single bonds with three Cl atoms. There a... | Chemistry 11th |
Explain the shapes of $\mathrm{NH}_{3}$ and $\mathrm{H}_{2}\mathrm{O}$ molecules using VSEPR theory. | For $$\mathrm{NH}_{3}$$, the central Nitrogen atom has 5 valence electrons. It forms 3 single bonds with Hydrogen atoms and has 1 lone pair of electrons, resulting in 4 electron domains. The 4 electron domains around Nitrogen adopt a tetrahedral electron geometry. Due to the one lone pair, the molecular geometry (the a... | Chemistry 11th |
What is valence bond theory? State its main features. | A covalent bond is formed by the overlapping of half-filled valence atomic orbitals of two different atoms. The electrons in the overlapping orbitals must have opposite spins. The strength of the bond depends on the extent of the overlap; greater overlap leads to a stronger bond. The geometry of a molecule is determine... | Chemistry 11th |
Explain the formation of $\mathrm{H}_{2}$ molecule using valence bond theory. | Consider two hydrogen atoms, each with one electron in a 1s orbital. When they are far apart, the potential energy of the system is zero. As the atoms approach each other, the 1s orbital of one atom begins to overlap with the 1s orbital of the other atom. A stable covalent bond forms when the two electrons, with opposi... | Chemistry 11th |
What is orbital overlap concept? Explain with the help of formation of hydrogen molecule. | According to valence bond theory, a covalent bond is formed when atomic orbitals of two different atoms merge or overlap. The strength of the bond depends on the extent of this overlap. For a stable bond to form, the overlapping orbitals must contain electrons with opposite spins. This allows the electrons to be shared... | Chemistry 11th |
What is positive overlap, negative overlap and zero overlap of orbitals? | Positive overlap occurs when the lobes of atomic orbitals with the same phase (e.g., both positive or both negative) interact. This constructive interference increases electron density between the nuclei, forming a stable bonding molecular orbital. Negative overlap occurs when the lobes of atomic orbitals with opposite... | Chemistry 11th |
Distinguish between sigma ($\sigma$) and pi ($\pi$) bonds. | Sigma ($\sigma$) bonds are formed by the direct, head-on overlap of atomic orbitals (like s-s, s-p, or p-p). Pi ($\pi$) bonds are formed by the sideways or lateral overlap of parallel p orbitals. In a sigma bond, the electron density is highest along the axis connecting the two nuclei. In a pi bond, the electron densit... | Chemistry 11th |
Explain s-s, s-p and p-p overlapping with diagrams. | s-s Overlap: This occurs when two spherical s-orbitals overlap along the internuclear axis. This head-on overlap forms a sigma ($$\sigma$$) bond. An example is the formation of the hydrogen molecule, $$H_2$$. s-p Overlap: This involves the overlap of a spherical s-orbital of one atom with a dumbbell-shaped p-orbital of... | Chemistry 11th |
What is hybridisation? What are the salient features of hybridisation? | Hybridisation is the concept of mixing atomic orbitals of slightly different energies to form a new set of orbitals with equivalent energies and shapes. These new orbitals are called hybrid orbitals. The number of hybrid orbitals formed is equal to the number of atomic orbitals that are mixed. Hybrid orbitals are alway... | Chemistry 11th |
What are the important conditions for hybridisation? | The orbitals undergoing hybridization should belong to the valence shell of the atom. The orbitals involved must have only a small difference in their energy levels. For example, 2s and 2p orbitals can hybridize, but 2s and 3p orbitals cannot. The number of hybrid orbitals formed is equal to the number of atomic orbita... | Chemistry 11th |
Explain $sp$ hybridisation with the example of $\mathrm{BeCl}_{2}$. | The central atom is Beryllium (Be), which is bonded to two Chlorine (Cl) atoms. Hybridization is the process of mixing atomic orbitals. In sp hybridization, one s orbital and one p orbital combine to form two new, identical sp hybrid orbitals. These orbitals arrange themselves linearly at an angle of 180 degrees. The g... | Chemistry 11th |
Explain $sp^{2}$ hybridisation with the example of $\mathrm{BCl}_{3}$. | The central atom is Boron ($$B$$), with atomic number 5. Its ground-state electron configuration is $$1s^2 2s^2 2p^1$$. To form three bonds with Chlorine, one electron from the $$2s$$ orbital is promoted to an empty $$2p$$ orbital. The excited-state configuration becomes $$1s^2 2s^1 2p_x^1 2p_y^1$$. The one $$2s$$ orbi... | Chemistry 11th |
Explain $sp^{3}$ hybridisation with the example of $\mathrm{CH}_{4}$. | The ground state electronic configuration of carbon (atomic number 6) is $$1s^2 2s^2 2p^2$$. It has only two unpaired electrons in the $$2p$$ orbital. To form four bonds, a $$2s$$ electron is promoted to the empty $$2p_z$$ orbital. The excited state configuration is $$1s^2 2s^1 2p_x^1 2p_y^1 2p_z^1$$, providing four un... | Chemistry 11th |
Explain $sp^{3}d$ hybridisation with the example of $\mathrm{PCl}_{5}$. | The central atom is Phosphorus ($$P$$). The atomic number of $$P$$ is 15. Its ground state electronic configuration is $$[Ne] 3s^{2}3p^{3}$$. It has only three unpaired electrons, but it needs to form five bonds with chlorine. To form five bonds, one electron from the $$3s$$ orbital is promoted to an empty $$3d$$ orbit... | Chemistry 11th |
Describe the bonding in ethane ($\mathrm{C}_{2}\mathrm{H}_{6}$) molecule. | Each carbon atom has four valence electrons and requires four bonds to be stable. To achieve this, it undergoes $$sp^3$$ hybridization. One $$2s$$ orbital and three $$2p$$ orbitals of each carbon atom mix to form four new, equivalent $$sp^3$$ hybrid orbitals. These orbitals arrange themselves in a tetrahedral geometry.... | Chemistry 11th |
Describe the bonding in ethyne ($\mathrm{C}_{2}\mathrm{H}_{2}$) molecule. | Each carbon atom in ethyne forms two sigma bonds (one with H, one with the other C) and has no lone pairs. This requires two hybrid orbitals, so each carbon is $$sp$$ hybridized. The remaining two $$p$$ orbitals on each carbon do not participate in hybridization. The C-C sigma ($$\sigma$$) bond is formed by the head-on... | Chemistry 11th |
What is molecular orbital theory? State its main features. | Atomic orbitals of comparable energy and proper symmetry combine to form an equal number of molecular orbitals. Molecular orbitals are of two types: bonding molecular orbitals (BMOs), which have lower energy and greater stability, and antibonding molecular orbitals (AMOs), which have higher energy and lower stability. ... | Chemistry 11th |
What is LCAO method? How are bonding and antibonding molecular orbitals formed? | The Linear Combination of Atomic Orbitals (LCAO) is a method to approximate molecular orbitals by combining atomic orbitals. Bonding molecular orbitals are formed by the constructive interference (addition) of atomic orbitals. This overlap increases electron density between the nuclei and results in a lower energy stat... | Chemistry 11th |
What are the conditions for effective combination of atomic orbitals to form molecular orbitals? | The combining atomic orbitals must have similar or comparable energy. A large energy difference prevents effective mixing. The combining atomic orbitals must have the same symmetry with respect to the molecular axis. For example, a $$p_z$$ orbital can combine with another $$p_z$$ orbital, but not with a $$p_x$$ orbital... | Chemistry 11th |
Distinguish between bonding and antibonding molecular orbitals. | Bonding molecular orbitals (MOs) result from the constructive interference (in-phase addition) of atomic orbitals, leading to increased electron density between the nuclei. Antibonding MOs result from the destructive interference (out-of-phase subtraction) of atomic orbitals, creating a node (zero electron density) bet... | Chemistry 11th |
What are sigma ($\sigma$) and pi ($\pi$) molecular orbitals? | Sigma ($\sigma$) orbitals result from the direct, head-on overlap of atomic orbitals (like s-s, s-p, or p-p). The electron density in a sigma bond is concentrated on the axis directly connecting the nuclei of the bonding atoms. Pi ($\pi$) orbitals result from the sideways or parallel overlap of p-orbitals. The electron... | Chemistry 11th |
Why is the energy order of molecular orbitals different for $\mathrm{O}_{2}$ and $\mathrm{N}_{2}$? | The ordering of molecular orbitals in diatomic molecules depends on the interaction between the 2s and 2p atomic orbitals. This interaction, called s-p mixing, occurs when the 2s and 2p orbitals are close in energy. It raises the energy of the $$\sigma_{2p}$$ orbital and lowers the energy of the $$\sigma_{2s}$$ orbital... | Chemistry 11th |
Write the molecular orbital configuration of $\mathrm{H}_{2}$ molecule and calculate its bond order. | The $$\mathrm{H}_{2}$$ molecule is formed from two hydrogen atoms, each contributing one electron. The total number of electrons is 2. The electrons fill the molecular orbitals starting with the lowest energy level. The resulting molecular orbital configuration is $$(\sigma_{1s})^2$$. The bond order is calculated using... | Chemistry 11th |
Explain why $\mathrm{He}_{2}$ molecule does not exist using molecular orbital theory. | A Helium atom ($$\mathrm{He}$$) has the electron configuration $$1s^2$$. A hypothetical $$\mathrm{He}_{2}$$ molecule would have a total of 4 electrons. The stability of a molecule is determined by its bond order. Bond Order = $$\frac{1}{2}$$ (Number of bonding electrons - Number of antibonding electrons). The 4 electro... | Chemistry 11th |
Write the molecular orbital configuration of $\mathrm{O}_{2}$ molecule and explain its paramagnetic nature. | An oxygen atom has 8 electrons, so the O2 molecule has a total of 16 electrons. The electrons are filled into molecular orbitals according to increasing energy levels: $$(\sigma1s), (\sigma^*1s), (\sigma2s), (\sigma^*2s), (\sigma2p_z), (\pi2p_x = \pi2p_y), (\pi^*2p_x = \pi^*2p_y), (\sigma^*2p_z)$$. Filling the 16 elect... | Chemistry 11th |
What are the types of hydrogen bonds? Explain with examples. | A hydrogen bond is an electrostatic attraction between a hydrogen atom covalently bonded to a highly electronegative atom (like N, O, or F) and another nearby electronegative atom. Intermolecular hydrogen bonding occurs between two or more different molecules. For example, the hydrogen bonds between water ($${H_2O}$$) ... | Chemistry 11th |
Give examples of intermolecular and intramolecular hydrogen bonding. | Intermolecular hydrogen bonding is an attractive force that occurs between two or more separate molecules. An example is the bonding between water molecules ($$\mathrm{H}_{2}\mathrm{O}$$). The partially positive hydrogen of one water molecule is attracted to the partially negative oxygen of a neighboring water molecule... | Chemistry 11th |
What is the effect of hydrogen bonding on the physical properties of compounds? | Hydrogen bonding is a strong type of intermolecular force that occurs when hydrogen is bonded to a highly electronegative atom like nitrogen (N), oxygen (O), or fluorine (F). Boiling and Melting Points: Compounds with hydrogen bonds have significantly higher boiling and melting points because more energy is required to... | Chemistry 11th |
Explain using $\Delta_G$ = $\Delta_H$ - T Delta S why some endothermic reactions are spontaneous at high temperatures. | For a spontaneous reaction, the change in Gibbs free energy ($$\Delta G$$) must be negative. For an endothermic reaction, the change in enthalpy ($$\Delta H$$) is positive. The Gibbs free energy equation is $$\Delta G = \Delta H - T\Delta S$$, where $$T$$ is temperature and $$\Delta S$$ is the change in entropy. For $$... | Chemistry 11th |
What is the relationship between standard Gibbs energy change and equilibrium constant? Write the equation. | The standard Gibbs energy change ($$\Delta G^\circ$$) is the change in Gibbs energy for a process when all reactants and products are in their standard states. The equilibrium constant ($$K$$) relates the concentrations or partial pressures of products and reactants at equilibrium. The relationship between them is give... | Chemistry 11th |
If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1 bar and 100 degrees C is 41 kJ/mol. Calculate the internal energy change, when 1 mol of water is vapourised at 1 bar pressure and 100 degrees C. | Given: Molar enthalpy change, $\Delta H = 41 \text{ kJ/mol}$. Pressure, $P = 1 \text{ bar}$. Temperature, $T = 100^\circ\text{C} = 373.15 \text{ K}$. Gas constant, $R = 8.314 \text{ J/(K} \cdot \text{mol)}$. The relationship between enthalpy change and internal energy change is given by the formula:
$\Delta H = \Delta... | Chemistry 11th |
A swimmer coming out from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporisation at 298 K. $\Delta_{vap} H^{0}$
for water at 298 K = 44.01 kJ/mol. | First, we find the number of moles of water. The molar mass of water ($\text{H}_2\text{O}$) is 18.015 g/mol.
$n = \frac{\text{mass}}{\text{molar mass}} = \frac{18 \text{ g}}{18.015 \text{ g/mol}} \approx 1 \text{ mol}$ The heat ($q$) required for evaporation is the product of the moles of water and the molar enthalpy ... | Chemistry 11th |
Express the change in internal energy of a system when\
(i) No heat is absorbed by the system from the surroundings, but work
(w) is done on the system. What type of wall does the system have?\
(ii) No work is done on the system, but q amount of heat is taken out from the system and given to the surroundings. What type... | For case (i), no heat is absorbed, so $$q = 0$$. Work is done on the system, so work ($$w$$) is positive. Since there is no heat exchange, the system must have an adiabatic wall. The change in internal energy ($$\Delta U$$) is found using the First Law of Thermodynamics, $$\Delta U = q + w$$. Substituting the values gi... | Chemistry 11th |
Choose the correct answer. A thermodynamic state function is a quantity\
(i) used to determine heat changes\
(ii) whose value is independent of path\
(iii) used to determine pressure volume work\
(iv) whose value depends on temperature only. | A state function is a property of a thermodynamic system that depends only on its current equilibrium state. The value of a state function does not depend on the process or path taken to reach that state. Heat ($$q$$) and work ($$w$$) are path functions, meaning their values depend on the specific path taken. Therefore... | Chemistry 11th |
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are -890.3 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol respectively. Enthalpy of formation of CH4
(g) will be\
(i) -74.8 kJ/mol\
(ii) -52.27 kJ/mol\
(iii) +74.8 kJ/mol\
(iv) +52.26 kJ/mol | The target reaction for the formation of methane is: $$C(s) + 2H_2(g) \rightarrow CH_4(g)$$. The given combustion reactions are:
1. $$CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l); \Delta H_c = -890.3 \text{ kJ/mol}$$
2. $$C(s) + O_2(g) \rightarrow CO_2(g); \Delta H_c = -393.5 \text{ kJ/mol}$$
3. $$H_2(g) + \frac{1}... | Chemistry 11th |
For the reaction
$$
2\,\mathrm{Cl\
(g)} \rightarrow \mathrm{Cl_2\
(g)},
$$
what are the signs of the enthalpy change, $\Delta H$, and the entropy change, $\Delta S$? | The reaction involves the formation of a Cl-Cl bond. Bond formation is an exothermic process, meaning it releases energy. For an exothermic reaction, the change in enthalpy ($$\Delta H$$) is negative. The reaction converts 2 moles of gaseous atoms into 1 mole of gaseous molecules. This decrease in the number of moles o... | Chemistry 11th |
What type of system would represent a chemical reaction occurring in a sealed copper container? What exchanges are possible? | The system is a chemical reaction inside a container. The container is described as 'sealed', which means it does not allow matter to pass in or out. The container is made of copper, which is a good conductor of heat. This allows energy to be exchanged between the system (the reaction) and the surroundings. A system th... | Chemistry 11th |
Explain why we cannot specify an absolute value for internal energy but can measure only changes in internal energy. | Internal energy ($$U$$) is the sum of all microscopic kinetic and potential energies of the particles in a system. It is impossible to measure the exact position, motion, and interaction of every single particle in a macroscopic system, so the absolute value of $$U$$ cannot be determined. The First Law of Thermodynamic... | Chemistry 11th |
If water vapour is assumed to be a perfect gas, calculate $\Delta_U$ when 1 mol of water is vaporized at 1 bar and 100 degrees C given $\Delta_{vap}$H = 41 kJ/mol. | Given: $$n=1$$ mol, $$P=1$$ bar, $$T = 100^\circ\text{C} = 373.15$$ K, and $$\Delta_{vap}H = 41$$ kJ/mol. The relationship between enthalpy and internal energy is $$\Delta H = \Delta U + \Delta n_g RT$$. We rearrange for $$\Delta U$$: $$\Delta U = \Delta H - \Delta n_g RT$$. For the vaporization $$\mathrm{H_2O}(l) \rig... | Chemistry 11th |
Describe how $\Delta_U$ is measured using a bomb calorimeter. Why is it called a constant volume process? | A substance is combusted in a sealed, rigid container called a 'bomb', which is submerged in a known quantity of water. The heat released by the combustion reaction ($$q_{rxn}$$) is absorbed by the bomb and the surrounding water, causing the temperature to rise. The total heat absorbed by the calorimeter ($$q_{cal}$$) ... | Chemistry 11th |
How can reaction enthalpy be estimated using bond enthalpies? What is the limitation of this method? | A chemical reaction involves breaking existing chemical bonds in the reactants and forming new chemical bonds in the products. The enthalpy of reaction ($$\Delta H$$) is the difference between the energy required to break bonds and the energy released when forming bonds. The formula is: $$\Delta H = \sum (\text{Bond en... | Chemistry 11th |
For oxidation of iron Delta S = -549.4 J/K/mol and $\Delta_r$ H = -1648 kJ/mol at 298 K. Explain why this reaction is spontaneous despite negative entropy change. | Given: Enthalpy change, $$\Delta H = -1648 \text{ kJ/mol}$$. Entropy change, $$\Delta S = -549.4 \text{ J/K/mol}$$. Temperature, $$T = 298 \text{ K}$$. The spontaneity of a reaction is determined by the Gibbs Free Energy change, using the formula: $$\Delta G = \Delta H - T\Delta S$$. A reaction is spontaneous if $$\Del... | Chemistry 11th |
Two litres of an ideal gas at a pressure of 10 atm expands isothermally at 25 degrees C into a vacuum until its total volume is 10 litres. How much heat is absorbed and how much work is done in the expansion?\
(i) Consider the same expansion as in Problem 5.2 but this time against a constant external pressure of 1 atm.... | **Part (i): Expansion into a Vacuum** The external pressure $$P_{ext}$$ is 0 for expansion into a vacuum. The formula for work done is $$w = -P_{ext} \Delta V$$. $$w = -(0 \text{ atm}) \times (10 \text{ L} - 2 \text{ L}) = 0$$. For an isothermal expansion of an ideal gas, the change in internal energy $$\Delta U = 0$$.... | Chemistry 11th |
How is entropy change related to heat and temperature? Write the expression for Delta S. | The change in entropy ($$\Delta S$$) of a system is a measure of the change in its disorder or randomness. Entropy change is directly proportional to the amount of heat ($$Q$$) added or removed from the system during a reversible process. It is inversely proportional to the absolute temperature ($$T$$) in Kelvin at whi... | Chemistry 11th |
Define Gibbs energy. Write the Gibbs equation and explain its significance. | Gibbs free energy ($$G$$) is the amount of energy available in a system to do useful work at constant temperature and pressure. The Gibbs free energy equation is: $$G = H - TS$$. Here, $$H$$ is enthalpy, $$T$$ is the absolute temperature in Kelvin, and $$S$$ is the entropy. For a process, the change in Gibbs free energ... | Chemistry 11th |
1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation C(graphite) + O2
(g) -> CO2
(g). During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the abo... | The heat absorbed by the calorimeter ($$q_{cal}$$) is calculated from the heat capacity ($$C_{cal}$$) and the temperature change ($$\Delta T$$). The heat released by the reaction at constant volume is the change in internal energy, $$\Delta U = -q_{cal}$$. The formula for the heat absorbed by the calorimeter is $$q_{ca... | Chemistry 11th |
Assuming the water vapour to be a perfect gas, calculate the internal energy change when 1 mol of water at 100 degrees C and 1 bar pressure is converted to ice at 0 degrees C. Given the enthalpy of fusion of ice is 6.00 kJ/mol, heat capacity of water is 4.2 J/g degrees C. | The total process is analyzed in three sequential steps: 1) Condensation of 1 mol of water vapor to liquid at 100 degrees C. 2) Cooling of 1 mol of liquid water from 100 degrees C to 0 degrees C. 3) Freezing of 1 mol of liquid water to ice at 0 degrees C. For the condensation of gas to liquid, the internal energy chang... | Chemistry 11th |
The combustion of one mole of benzene takes place at 298 K and 1 atm. After combustion, CO2
(g) and H2O
(l) are produced and 3267.0 kJ of heat is liberated. Calculate the standard enthalpy of formation of benzene. Standard enthalpies of formation of CO2
(g) and H2O
(l) are -393.5 kJ/mol and -285.83 kJ/mol respectively. | First, write the balanced thermochemical equation for the combustion of benzene:
$\text{C}_6\text{H}_6\text{(l)} + \frac{15}{2}\text{O}_2\text{(g)} \rightarrow 6\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)}$
with $\Delta H_c^\circ = -3267.0 \text{ kJ/mol}$. The standard enthalpy of a reaction is calculated using the... | Chemistry 11th |
Predict in which of the following, entropy increases/decreases:\
(i) A liquid crystallizes into a solid.\
(ii) Temperature of a crystalline solid is raised from 0 K to 115 K.\
(iii) 2NaHCO3
(s) -> Na2CO3
(s) + CO2
(g) + H2O
(g)
(iv) H2
(g) -> 2H
(g) | (i) A liquid crystallizes into a solid: Particles in a liquid are in a disordered state, while in a solid they are in a fixed, ordered arrangement. The transition from a disordered state to an ordered state causes a decrease in entropy. (ii) Temperature of a crystalline solid is raised from 0 K to 115 K: As the tempera... | Chemistry 11th |
For oxidation of iron, 4Fe
(s) + 3O2
(g) -> 2Fe2O3
(s), entropy change is -549.4 J/K/mol at 298 K. Inspite of negative entropy change of this reaction, why is the reaction spontaneous? ($\Delta_r H^{0}$
for this reaction is -1648 x $10^3$ J/mol) | Given: $$\Delta S = -549.4 \text{ J/K/mol}$$, $$\Delta H = -1648 \times 10^3 \text{ J/mol}$$, and $$T = 298 \text{ K}$$. The spontaneity of a reaction is determined by the Gibbs Free Energy equation: $$\Delta G = \Delta H - T\Delta S$$. A reaction is spontaneous if $$\Delta G$$ is negative. Substituting the values: $$\... | Chemistry 11th |
Calculate the, $\Delta_r G^\circ$, for the conversion of oxygen to ozone:
$\frac{3}{2}\,\mathrm{O_2
(g)} \rightarrow \mathrm{O_3
(g)}$ at $298\,\text{K}$, if the equilibrium constant $K_p$ for this conversion is
\[
$K_p = 2.47 \times 10^{-29}.$
\] | The given values are the equilibrium constant $$K_p = 2.47 \times 10^{-29}$$ and the temperature $$T = 298\,\text{K}$$. The ideal gas constant is $$R = 8.314\,\text{J K}^{-1}\,\text{mol}^{-1}$$. The formula relating the standard Gibbs free energy change to the equilibrium constant is: $$\Delta_r G^\circ = -RT \ln K_p$$... | Chemistry 11th |
Find out the value of equilibrium constant for the following reaction at 298 K: 2NH3
(g) + CO2
(g) <=> NH2CONH2(aq) + H2O
(l). Standard Gibbs energy change at the given temperature is -13.6 kJ/mol. | The given standard Gibbs energy change is $$\Delta G^\circ = -13.6 \ \text{kJ/mol}$$, which is equal to $$-13600 \ \text{J/mol}$$. The temperature is $$T = 298 \ \text{K}$$, and the universal gas constant is $$R = 8.314 \ \text{J K}^{-1} \text{mol}^{-1}$$. The relationship between the standard Gibbs free energy change ... | Chemistry 11th |
At 60 degrees C, dinitrogen tetroxide is 50 per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere. | The reaction is $\text{N}_2\text{O}_4\text{(g)} \rightleftharpoons 2\text{NO}_2\text{(g)}$. The given data are: Temperature $T = 60^\circ\text{C} = 333.15 \text{ K}$, total pressure $P = 1 \text{ atm}$, and degree of dissociation $\alpha = 0.50$. The equilibrium constant in terms of partial pressure, $K_p$, is related ... | Chemistry 11th |
For the process to occur under adiabatic conditions, the correct condition is:
$$
\begin{aligned}
\text{\
(i)}\;& \Delta T = 0 \
\text{\
(ii)}\;& \Delta p = 0 \
\text{\
(iii)}\;& q = 0 \
\text{\
(iv)}\;& w = 0
\end{aligned}
$$ | An adiabatic process is a thermodynamic process where no heat is exchanged between the system and its surroundings. The variable used to represent heat transfer is $$q$$. For a process with no heat transfer, the value of $$q$$ must be zero. Therefore, the correct condition for an adiabatic process is $$q = 0$$. **Final... | Chemistry 11th |
The enthalpies of all elements in their standard states are:\
(i) unity\
(ii) zero\
(iii) < 0\
(iv) different for each element | The problem asks for the value of the standard enthalpy of formation ($$\Delta H_f^\circ$$) for elements in their standard states. By definition, the standard enthalpy of formation of an element in its most stable form (standard state) is set as the reference point. This reference point is defined as zero. For example,... | Chemistry 11th |
$$
\Delta U = -X \,\text{kJ mol}^{-1}.
$$
The value of enthalpy change, $\Delta H$, is:
$$
\begin{aligned}
\text{\
(i)}\;& \Delta H = \Delta U \
\text{\
(ii)}\;& \Delta H > \Delta U \
\text{\
(iii)}\;& \Delta H < \Delta U \
\text{\
(iv)}\;& \Delta H = 0
\end{aligned}
$$ | The given reaction is $$\mathrm{NH_2CN(s) + \frac{3}{2}O_2(g) \rightarrow N_2(g) + CO_2(g) + H_2O(l)}$$. The change in internal energy is $$\Delta U = -X \,\text{kJ mol}^{-1}$$. The relationship between enthalpy and internal energy is given by the formula: $$\Delta H = \Delta U + \Delta n_g RT$$. Here, $$\Delta n_g$$ i... | Chemistry 11th |
A reaction, A + B -> C + D + q is found to have a positive entropy change. The reaction will be\
(i) possible at high temperature\
(ii) possible only at low temperature\
(iii) not possible at any temperature\
(iv) possible at any temperature | The reaction A + B -> C + D + q releases heat, so it is exothermic. This means the change in enthalpy ($$\Delta H$$) is negative. The problem states that the entropy change ($$\Delta S$$) is positive. The spontaneity of a reaction is determined by the Gibbs free energy equation: $$\Delta G = \Delta H - T\Delta S$$. Sub... | Chemistry 11th |
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process? | Heat absorbed by the system ($$q$$) is +701 J. Work done by the system ($$w$$) is -394 J. The First Law of Thermodynamics is given by the equation: $$\Delta U = q + w$$. Substitute the given values into the formula: $$\Delta U = (701 \text{ J}) + (-394 \text{ J})$$. This calculates to $$\Delta U = 307 \text{ J}$$. The ... | Chemistry 11th |
The reaction of cyanamide, NH2CN
(s), with dioxygen was carried out in a bomb calorimeter, and $$\Delta U$$ was found to be -742.7 kJ/mol at 298 K. Calculate enthalpy change for the reaction at 298 K. NH2CN
(g) + 3/2 O2
(g) -> N2
(g) + CO2
(g) + H2O
(l) | Given: The internal energy change, $\Delta U = -742.7 ext{ kJ/mol}$. The temperature, $T = 298 ext{ K}$. The ideal gas constant, $R = 8.314 ext{ J/(mol} \cdot ext{K)}$. The reaction is:
$ ext{NH}_2 ext{CN(s)} + rac{3}{2} ext{O}_2 ext{(g)}
ightarrow ext{N}_2 ext{(g)} + ext{CO}_2 ext{(g)} + ext{H}_2 ext{O(l)}$... | Chemistry 11th |
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35 degrees C to 55 degrees C. Molar heat capacity of Al is 24 J/mol/K. | Given: mass of Al = 60.0 g, initial temperature $T_i = 35^\circ\text{C}$, final temperature $T_f = 55^\circ\text{C}$, and molar heat capacity $C_m = 24 \text{ J/(mol} \cdot \text{K)}$. The molar mass of Al is 26.98 g/mol. The formula to calculate heat is $q = nC_m\Delta T$, where $n$ is the number of moles and $\Delta ... | Chemistry 11th |
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0 degrees C to ice at -10.0 degrees C. $\Delta_{fus}$ H = 6.03 kJ/mol at 0 degrees C. Cp[H2O
(l)] = 75.3 J/mol/K, Cp[H2O
(s)] = 36.8 J/mol/K | The process is broken into three steps: cooling the liquid water, freezing the water into ice, and cooling the ice. The total enthalpy change is the sum of the changes in each step:
$\Delta H_{total} = \Delta H_{cooling, liquid} + \Delta H_{freezing} + \Delta H_{cooling, solid}$ Step 1: Enthalpy change for cooling liq... | Chemistry 11th |
Enthalpy of combustion of carbon to CO2 is -393.5 kJ/mol. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and dioxygen gas. | The enthalpy of combustion of carbon to CO2 is -393.5 kJ/mol. The mass of CO2 is 35.2 g. The formula to find the moles (n) is $$n = \frac{\text{mass}}{\text{Molar Mass}}$$. The formula for heat released (q) is $$q = n \times \Delta H_{\text{combustion}}$$. The molar mass of CO2 is $$12.01 + 2(16.00) = 44.01 \text{ g/mo... | Chemistry 11th |
Enthalpies of formation of CO
(g), CO2
(g), N2O
(g) and N2O4
(g) are -110, -393, 81 and 9.7 kJ/mol respectively. Find the value of
$$\Delta_r H$$ for the reaction: N2O4
(g) + 3CO
(g) -> N2O
(g) + 3CO2
(g) | The given standard enthalpies of formation ($$\Delta_f H$$) are: $$\Delta_f H(\mathrm{CO})$$ = -110 kJ/mol, $$\Delta_f H(\mathrm{CO_2})$$ = -393 kJ/mol, $$\Delta_f H(\mathrm{N_2O})$$ = 81 kJ/mol, and $$\Delta_f H(\mathrm{N_2O_4})$$ = 9.7 kJ/mol. The formula for the standard enthalpy of reaction is: $$\Delta_r H = \sum ... | Chemistry 11th |
Given N2
(g) + 3H2
(g) -> 2NH3
(g); $\Delta_r$ H = -92.4 kJ/mol. What is the standard enthalpy of formation of NH3 gas? | The given reaction shows the formation of 2 moles of ammonia ($$\mathrm{NH_3}$$) from its elements in their standard states. The enthalpy change for this reaction is $$\Delta_r H$$ = -92.4 kJ/mol. The standard enthalpy of formation ($$\Delta_f H^\circ$$) is the enthalpy change for the formation of 1 mole of a substance... | Chemistry 11th |
Calculate the standard enthalpy of formation of CH3OH
(l) from the following data: CH3OH
(l) + 3/2 O2
(g) -> CO2
(g) + 2H2O
(l); Delta_r H = -726 kJ/mol. C(graphite) + O2
(g) -> CO2
(g); Delta_c H = -393 kJ/mol. H2
(g) + 1/2 O2
(g) -> H2O
(l); Delta_f H = -286 kJ/mol. | The target reaction is the formation of $$\mathrm{CH_3OH(l)}$$ from its elements in their standard states: C(graphite) + 2H2(g) + 1/2 O2(g) -> CH3OH(l). The given data is:
(i) $$\mathrm{CH_3OH(l) + \frac{3}{2}O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}$$; $$\Delta H = -726$$ kJ/mol
(ii) $$\mathrm{C(graphite) + O_2(g) \right... | Chemistry 11th |
Calculate the enthalpy change, $\Delta_r H$, for the process $$\mathrm{CCl_4\
(g)} \rightarrow \mathrm{C\
(g)} + 4\,\mathrm{Cl\
(g)}$$ and hence calculate the bond enthalpy of the $\mathrm{C{-}Cl}$ bond in $\mathrm{CCl_4\
(g)}$.\
Given data:$$\Delta_{\mathrm{vap}} H(\mathrm{CCl_4}) = 30.5 \,\text{kJ mol}^{-1}$$
$$
\Del... | Given Data:
Enthalpy of vaporization, $\Delta_{\text{vap}} H(\text{CCl}_4) = 30.5 \text{ kJ mol}^{-1}$
Enthalpy of formation, $\Delta_f H^\circ(\text{CCl}_4\text{, l}) = -135.5 \text{ kJ mol}^{-1}$
Enthalpy of atomization of Carbon, $\Delta_a H(\text{C}) = 715.0 \text{ kJ mol}^{-1}$
Enthalpy of atomization of Chlor... | Chemistry 11th |
For an isolated system, $\Delta U = 0$. What will be the value of $\Delta S$? | An isolated system is one that does not exchange energy or matter with its surroundings. The Second Law of Thermodynamics states that for any spontaneous process occurring in an isolated system, the change in entropy must be greater than or equal to zero. For a spontaneous, irreversible process, the entropy increases. ... | Chemistry 11th |
For the reaction at $298\,\text{K}$,
$$
2A + B \rightarrow C
$$
the enthalpy change and entropy change are given as:
$$
\Delta H = 400 \,\text{kJ mol}^{-1}, \qquad
\Delta S = 0.2 \,\text{kJ K}^{-1}\text{mol}^{-1}.
$$
At what temperature will the reaction become spontaneous, assuming that
$\Delta H$ and $\Delta S$ rema... | Given enthalpy change, $$\Delta H = 400 \,\text{kJ mol}^{-1}$$. Given entropy change, $$\Delta S = 0.2 \,\text{kJ K}^{-1}\text{mol}^{-1}$$. The Gibbs free energy equation is $$\Delta G = \Delta H - T\Delta S$$. A reaction becomes spontaneous when $$\Delta G < 0$$. The transition temperature is found by setting $$\Delta... | Chemistry 11th |
For the reaction
$$
2A\
(g) + B\
(g) \rightarrow 2D\
(g),
$$
the change in internal energy and entropy are given as:
$$
\Delta U = -10.5 \,\text{kJ}, \qquad
\Delta S = -44.1 \,\text{J K}^{-1}.
$$
Calculate the Gibbs free energy change, $\Delta G$, for the reaction and predict whether the reaction may occur spontaneous... | Given data: $$\Delta U = -10.5$$ kJ and $$\Delta S = -44.1$$ J/K. We assume a standard temperature of $$T = 298$$ K. First, relate enthalpy ($$\Delta H$$) and internal energy ($$\Delta U$$) with the formula $$\Delta H = \Delta U + \Delta n_{gas}RT$$. Then, use the Gibbs free energy equation, $$\Delta G = \Delta H - T\D... | Chemistry 11th |
The equilibrium constant for a reaction is $K = 10$.
Calculate the Gibbs free energy change, $\Delta G$, at
$$
T = 300\,\text{K},
$$
given that
$$
R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}.
$$ | Given: Equilibrium constant $$K = 10$$, Temperature $$T = 300$$ K, and Gas Constant $$R = 8.314$$ J K$$^{-1}$$mol$$^{-1}$$. The formula relating Gibbs free energy, temperature, and the equilibrium constant is: $$\Delta G = -RT \ln K$$ Substitute the values into the equation: $$\Delta G = -(8.314\,\text{J K}^{-1}\text{m... | Chemistry 11th |
Comment on the thermodynamic stability of $\mathrm{NO\
(g)}$, given the following reactions:
$$
\frac{1}{2}\,\mathrm{N_2\
(g)} + \frac{1}{2}\,\mathrm{O_2\
(g)} \rightarrow \mathrm{NO\
(g)},
\qquad \Delta_r H = 90 \,\text{kJ mol}^{-1}
$$
$$
\mathrm{NO\
(g)} + \frac{1}{2}\,\mathrm{O_2\
(g)} \rightarrow \mathrm{NO_2\
(... | The formation of $$\mathrm{NO(g)}$$ from its elements is an endothermic reaction, as shown by the positive enthalpy change. $$\frac{1}{2}\,\mathrm{N_2\ (g)} + \frac{1}{2}\,\mathrm{O_2\ (g)} \rightarrow \mathrm{NO\ (g)}, \qquad \Delta_r H = +90 \,\text{kJ mol}^{-1}$$ Because energy is absorbed to form $$\mathrm{NO(g)}$$... | Chemistry 11th |
Calculate the entropy change of the surroundings, $\Delta S_{\text{surr}}$, when
$1.00\,\text{mol}$ of $\mathrm{H_2O\
(l)}$ is formed under standard conditions.
Given:
$$
\Delta_f H^\circ(\mathrm{H_2O\
(l)}) = -286 \,\text{kJ mol}^{-1}.
$$ | Given: Enthalpy of formation, $$\Delta H^\circ_f$$ = -286 kJ/mol. Moles, $$n$$ = 1.00 mol. Standard temperature, $$T$$ = 298 K. The formula for the entropy change of the surroundings is $$\Delta S_{\text{surr}} = \frac{-\Delta H_{\text{sys}}}{T}$$. First, find the total enthalpy change: $$\Delta H_{\text{sys}} = 1.00\,... | Chemistry 11th |
Define a thermodynamic system and its surroundings. What constitutes the boundary between them? | A thermodynamic system is the specific portion of the universe that is being studied. The surroundings are everything outside of the system. The surroundings are also called the environment. The boundary is the real or imaginary surface that separates the system from its surroundings. It controls the exchange of energy... | Chemistry 11th |
What are state functions? Give examples and explain why they are important in thermodynamics. | A state function is a property of a system that depends only on its current equilibrium state. It is independent of the path taken to reach that state. Common examples of state functions include Pressure ($$P$$), Volume ($$V$$), Temperature ($$T$$), Internal Energy ($$U$$), Enthalpy ($$H$$), and Entropy ($$S$$). In con... | Chemistry 11th |
State the first law of thermodynamics and express it mathematically. What does it signify? | The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed in an isolated system. The mathematical expression for the first law is: $$\Delta U = Q - W$$ In the equation, $$\Delta U$$ represents the change in the internal energy of the system, $$Q$... | Chemistry 11th |
What are the sign conventions for heat
(q) and work
(w) according to IUPAC? | All signs are determined from the perspective of the system. Energy entering the system is positive. Heat absorbed by the system is endothermic ($$q > 0$$). Work done on the system is also positive ($$w > 0$$). Energy leaving the system is negative. Heat released by the system is exothermic ($$q < 0$$). Work done by th... | Chemistry 11th |
How does internal energy of a system change? List the ways through which internal energy can be altered. | Internal energy ($$U$$) is the total kinetic and potential energy of all the particles within a system. The change in internal energy ($$\Delta U$$) is described by the First Law of Thermodynamics. The formula is $$ \Delta U = Q - W $$, where $$Q$$ is the heat added to the system and $$W$$ is the work done by the syste... | Chemistry 11th |
Distinguish between open, closed and isolated systems with examples. | An open system exchanges both energy and matter with its surroundings. An example is a boiling pot of water without a lid, which releases heat and water vapor. A closed system exchanges energy but not matter with its surroundings. An example is a sealed pressure cooker, which can be heated or cooled but does not lose i... | Chemistry 11th |
A thermos flask containing hot coffee is an example of which type of system? Justify your answer. | Thermodynamic systems are classified based on the exchange of energy and matter with their surroundings. An open system exchanges both, a closed system exchanges only energy, and an isolated system exchanges neither. A thermos flask is designed with a vacuum insulation layer to prevent heat transfer (energy exchange) b... | Chemistry 11th |
What is an adiabatic process? How is internal energy change related to work in such a process? | An adiabatic process is a thermodynamic process where no heat is transferred into or out of the system. This means the heat exchange, $$Q$$, is zero. The First Law of Thermodynamics relates internal energy ($$\Delta U$$), heat ($$Q$$), and work ($$W$$) as: $$\Delta U = Q - W$$. Substituting $$Q = 0$$ for an adiabatic p... | Chemistry 11th |
In a process 701 J of heat is absorbed by a system and 394 J of work is done by the system. Calculate the change in internal energy. | Heat absorbed by the system, $$q = +701$$ J. Work done by the system, $$w = -394$$ J. The formula is the First Law of Thermodynamics: $$\Delta U = q + w$$. Substitute the values: $$\Delta U = 701 + (-394)$$. This gives $$\Delta U = 307$$ J. The change in internal energy is 307 J. **Final Answer:** 307 J | Chemistry 11th |
Express the change in internal energy when:\
(i) no heat is absorbed but work is done on the system\
(ii) no work is done but heat is taken out from the system. | For case (i), no heat is absorbed, so $$q = 0$$. Work is done on the system, so $$w$$ is positive. This process is adiabatic. The First Law of Thermodynamics states $$\Delta U = q + w$$. Substituting the values for case (i): $$\Delta U = 0 + w$$, which simplifies to $$\Delta U = w$$. The internal energy increases. For ... | Chemistry 11th |
Two litres of an ideal gas at 10 atm expands isothermally at 25 degrees C into vacuum until volume is 10 litres. Calculate heat absorbed and work done. | The given data is: Initial Volume $$V_1 = 2$$ L, Final Volume $$V_2 = 10$$ L. The expansion occurs into a vacuum, so the external pressure $$P_{ext} = 0$$. The process is isothermal, meaning the temperature is constant. The formula for work done is $$w = -P_{ext} \Delta V$$. According to the First Law of Thermodynamics... | Chemistry 11th |
Derive the expression for work done during isothermal reversible expansion of n moles of an ideal gas. | The work done ($$W$$) by a gas during a reversible expansion from an initial volume $$V_1$$ to a final volume $$V_2$$ is given by the integral $$W = -\int_{V_1}^{V_2} P dV$$. For an ideal gas, the pressure $$P$$ can be expressed using the ideal gas law, $$PV = nRT$$. Rearranging for pressure gives $$P = \frac{nRT}{V}$$... | Chemistry 11th |
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