Update main.py

main.py

@@ -1,37 +1,97 @@
-from fastapi import FastAPI
-from groq import Groq
-
-app = FastAPI()
-
-client = Groq(
-    api_key="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",
-)
-
-@app.post("/chat")
-async def chat(query: str):
-    chat_completion = client.chat.completions.create(
-        messages=[
-            {
-                "role": "system",
-                "content": """Generate me 10 MCQ type questions on the topic asked by the user of Joint Entrance Examination level and also explain the solution of each problem step-wise. And also give a hardness rating out of 100 , The questions should have difficulty matching with the level of JEE mains and advanced.
-                Give me questions which are of very high difficulty level.
-
-                         Structure of output :
-                                        Question : "<Question String>"
-                                        Options : "<option 1, option 2, option 3, option 4>"
-                                        Answer : "<answer String>"
-                                        Hardness rating : "<hardness number>"
-                                        Solution : "<solution String>"
-
-                Make sure you provide VERY VERY Advanced level questions which cover the given topic properly , also the solutions or Hints has to be really relevant and explainatory.
-                """
-            },
-            {
-                "role": "user",
-                "content": query,
-            }
-        ],
-        model="llama3-70b-8192",
-    )
-
+from fastapi import FastAPI
+from groq import Groq
+
+app = FastAPI()
+
+client = Groq(
+    api_key="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",
+)
+
+@app.post("/chat")
+async def chat(query: str):
+    chat_completion = client.chat.completions.create(
+        messages=[
+            {
+                "role": "system",
+                "content": """Generate me 10 MCQ type questions on the topic asked by the user of Joint Entrance Examination level and also explain the solution of each problem step-wise. And also give a hardness rating out of 100 , The questions should have difficulty matching with the level of JEE mains and advanced.
+                Give me questions which are of very high difficulty level.
+                Make sure you provide VERY VERY Advanced level questions which cover the given topic properly , also the solutions or Hints has to be really relevant and explainatory.
+                                Expected output format:
+                {"response":"Here are 10 MCQ type questions on the topic of Equilibrium at the level of Joint 
+                Entrance Examination (JEE):\n\n**Question 1:**\nQuestion: A container having a fixed volume 
+                of 1 liter is filled with 1 mole of a gas. The temperature of the system is increased from 27°C to 
+                127°C. The entropy change of the system is (R = 8.314 J/mol-K)\nOptions: 23.4 J/mol-K, 46.8 
+                J/mol-K, 29.3 J/mol-K, 58.6 J/mol-K\nAnswer: 23.4 J/mol-K\nHardness rating: 80\nSolution: The 
+                entropy change of an ideal gas can be calculated using the formula ΔS = nR ln(V2/V1) + nCp 
+                ln(T2/T1). Since the volume is constant, the first term is zero. Plugging in the values, we get ΔS = 
+                1 Ô 8.314 Ô ln(400/300) = 23.4 J/mol-K.\n\n**Question 2:**\nQuestion: A gas phase reaction A 
+                └ B is in equilibrium at a certain temperature. If the initial concentration of A is 2M and the 
+                equilibrium concentration of A is 1.5M, what is the equilibrium constant (Kc) of the 
+                reaction?\nOptions: 0.25, 1/3, 0.5, 1/4\nAnswer: 0.5\nHardness rating: 70\nSolution: Let the 
+                initial concentration of A be a, and the equilibrium concentrations of A and B be a-x and x 
+                respectively. Then, Kc = [B]/[A] = x/(a-x). Plugging in the values, we get Kc = (0.5)/(1.5) = 
+                0.5.\n\n**Question 3:**\nQuestion: A 1L container is divided into two compartments of equal 
+                volume by a piston. One compartment contains 1 mole of an ideal gas at 300K, and the other 
+                compartment is evacuated. If the piston is removed and the system is allowed to reach 
+                equilibrium, what is the final temperature of the system? (R = 8.314 J/mol-K)\nOptions: 225 K, 
+                300 K, 450 K, 600 K\nAnswer: 300 K\nHardness rating: 90\nSolution: Since the system is 
+                isolated, the total internal energy remains constant. The initial energy of the system is (3/2)nRT = 
+                (3/2) Ô 1 Ô 8.314 Ô 300 = 3741 J. After the piston is removed, the gas expands to double its 
+                volume, and the final temperature can be calculated using the formula U = (3/2)nRT. Plugging in 
+                the values, we get T = 300 K.\n\n**Question 4:**\nQuestion: The equilibrium constant (Kp) for 
+                the reaction CO(g) + Cl2(g) └ COCl2(g) is 3.4 Ô 10^3 at 700K. If the initial pressures of CO and 
+                Cl2 are 0.1 atm and 0.2 atm respectively, what is the equilibrium pressure of COCl2?\nOptions: 
+                0.68 atm, 1.36 atm, 2.04 atm, 3.4 atm\nAnswer: 0.68 atm\nHardness rating: 85\nSolution: Let 
+                the initial pressures of CO, Cl2, and COCl2 be p, 2p, and 0 respectively. At equilibrium, the 
+                pressures are p-x, 2p-x, and x respectively. The equilibrium constant can be expressed as Kp = 
+                x/(p-x)(2p-x). Plugging in the values, we get x = 0.68 atm.\n\n**Question 5:**\nQuestion: The 
+                standard Gibbs free energy change (ΔG°) for a reaction is -20 kJ/mol at 298K. What is the 
+                equilibrium constant (K) for the reaction?\nOptions: 10^3, 10^4, 10^5, 10^6\nAnswer: 
+                10^4\nHardness rating: 75\nSolution: The equilibrium constant can be calculated using the 
+                formula ΔG° = -RT lnK. Plugging in the values, we get lnK = 20/(-8.314 Ô 298) = 8.14. Therefore, 
+                K = 10^4.\n\n**Question 6:**\nQuestion: A reaction A + B └ C + D has an equilibrium constant 
+                (Kc) of 4 at a certain temperature. If the initial concentrations of A, B, and C are 2M, 3M, and 1M 
+                respectively, what is the equilibrium concentration of D?\nOptions: 1M, 2M, 3M, 4M\nAnswer: 
+                2M\nHardness rating: 80\nSolution: Let the equilibrium concentrations of A, B, C, and D be a-x, 
+                b-x, c+x, and x respectively. The equilibrium constant can be expressed as Kc = (c+x)(x)/(a-x)(bx) = 4. Plugging in the values, we get x = 2M.\n\n**Question 7:**\nQuestion: The equilibrium
+                constant (Kp) for the reaction H2(g) + I2(g) └ 2HI(g) is 50 at a certain temperature. If the initial 
+                pressures of H2 and I2 are 1 atm and 2 atm respectively, what is the degree of dissociation of HI 
+                at equilibrium?\nOptions: 0.4, 0.6, 0.8, 0.9\nAnswer: 0.8\nHardness rating: 90\nSolution: Let 
+                the initial pressures of H2, I2, and HI be p, 2p, and 0 respectively. At equilibrium, the pressures 
+                are p-x, 2p-x, and 2x respectively. The equilibrium constant can be expressed as Kp = (2x)^2/(px)(2p-x) = 50. Plugging in the values, we get x = 0.8p. Therefore, the degree of dissociation is 
+                0.8.\n\n**Question 8:**\nQuestion: A reaction A └ B is in equilibrium at a certain 
+                temperature. If the entropy of A and B are 200 J/mol-K and 250 J/mol-K respectively, what is the 
+                equilibrium constant (K) of the reaction?\nOptions: 2, 4, 6, 8\nAnswer: 4\nHardness rating: 
+                80\nSolution: The equilibrium constant can be calculated using the formula ΔG° = -RT lnK = 
+                ΔH° - TΔS°. Since the reaction is at equilibrium, ΔG° = 0. Plugging in the values, we get 
+                ΔH° = TΔS°. Therefore, K = exp(ΔS°/R) = 4.\n\n**Question 9:**\nQuestion: The equilibrium 
+                constant (Kc) for the reaction A + 2B └ C is 10 at a certain temperature. If the initial 
+                concentrations of A and B are 1M and 2M respectively, what is the equilibrium concentration of 
+                C?\nOptions: 1M, 2M, 4M, 6M\nAnswer: 2M\nHardness rating: 85\nSolution: Let the equilibrium 
+                concentrations of A, B, and C be a-x, b-2x, and x respectively. The equilibrium constant can be 
+                expressed as Kc = x/(a-x)(b-2x)^2 = 10. Plugging in the values, we get x = 2M.\n\n**Question 
+                10:**\nQuestion: A 2L container is divided into two compartments of equal volume by a piston. 
+                One compartment contains 1 mole of an ideal gas at 300K, and the other compartment is 
+                evacuated. If the piston is removed and the system is allowed to reach equilibrium, what is the 
+                final entropy of the system? (R = 8.314 J/mol-K)\nOptions: 23.4 J/mol-K, 46.8 J/mol-K, 58.6 
+                J/mol-K, 69.4 J/mol-K\nAnswer: 46.8 J/mol-K\nHardness rating: 95\nSolution: Since the system 
+                is isolated, the total internal energy remains constant. The initial energy of the system is 
+                (3/2)nRT = (3/2) Ô 1 Ô 8.314 Ô 300 = 3741 J. After the piston is removed, the gas expands to 
+                double its volume, and the final entropy can be calculated using the formula ΔS = nR ln(V2/V1) 
+                + nCp ln(T2/T1). Since the temperature remains constant, the second term is zero. Plugging in 
+                the values, we get ΔS = 1 Ô 8.314 Ô ln(2) = 5.763 J/mol-K. Therefore, the final entropy is 23.4 + 
+                5.763 = 46.8 J/mol-K."}
+
+                VERY IMPORATANT : MAKE SURE YOU STRICTLT FOLLOW THE OUTPUT FORMAT .
+
+                """
+            },
+            {
+                "role": "user",
+                "content": query,
+            }
+        ],
+        model="llama3-70b-8192",
+    )
+
     return {"response": chat_completion.choices[0].message.content}