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Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.?
Verified Answer
Two rigid adiabatic vessel A and B which initially contain two gases a...
R = 0.082 atm/mole
n total = 6 (2 + 4)
V = 12 L (5 + 7)
Heat exchange at constant volume: n Cv delta(T)
2 x (3/2) R (Tf - 300) + 4 x (5/2) R (Tf - 400) = 5800
3R (Tf - 300) + 10R (Tf - 400) = 5800
13R Tf = 5800+9800
Tf = 975 K
PV=nRT
Pf V= n total x R x Tf
Pf = (6 x 0.08 x 975)/12
Pf= 39 atm
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Most Upvoted Answer
Two rigid adiabatic vessel A and B which initially contain two gases a...
Given:
- Two rigid adiabatic vessels A and B
- Vessel A contains 2 moles of He gas at 300 K
- Vessel B contains 4 moles of O2 gas at 400 K
- Cp,m (specific heat capacity at constant pressure) = 2.5 R
- Cv,m (specific heat capacity at constant volume) = 2.5 R
- Volume of vessel A = 5 liters
- Volume of vessel B = 7 liters
- Heat supplied = 5.8 kcal
To find:
- Final pressure in the vessels
Explanation:
1. First, let's calculate the initial pressures in the vessels using the ideal gas law:
- For vessel A: PV = nRT
- P_A * V_A = n_A * R * T_A
- P_A = (n_A * R * T_A) / V_A
- P_A = (2 * R * 300) / 5
- P_A = 120 R/5
- For vessel B: PV = nRT
- P_B * V_B = n_B * R * T_B
- P_B = (n_B * R * T_B) / V_B
- P_B = (4 * R * 400) / 7
- P_B = 1600 R/7
2. Now, let's calculate the change in internal energy for each vessel:
- For vessel A: ΔU_A = n_A * Cv,m * ΔT_A
- ΔU_A = 2 * 2.5R * (T_final - 300)
- For vessel B: ΔU_B = n_B * Cv,m * ΔT_B
- ΔU_B = 4 * 2.5R * (T_final - 400)
3. Since the vessels are connected by a pipe of negligible volume, the total moles of gas remain constant.
- Therefore, n_A + n_B = 2 + 4 = 6 moles
4. The total change in internal energy is the sum of the changes in each vessel:
- ΔU_total = ΔU_A + ΔU_B
- ΔU_total = (2 * 2.5R * (T_final - 300)) + (4 * 2.5R * (T_final - 400))
5. According to the first law of thermodynamics, ΔU_total = q + w, where q is heat supplied and w is work done. Since the vessels are adiabatic (no heat exchange with the surroundings), q = 0.
- Therefore, ΔU_total = w
6. The work done is given by w = -PΔV, where ΔV is the change in volume. The volumes of the vessels remain constant, so ΔV_A = ΔV_B = 0.
7. Therefore, w = 0, and ΔU_total = 0.
8. Plugging in the values, we get:
- (2 * 2.5R * (T_final - 300)) + (4 * 2.5R * (T_final - 400)) = 0
9. Solving the equation, we
- Two rigid adiabatic vessels A and B
- Vessel A contains 2 moles of He gas at 300 K
- Vessel B contains 4 moles of O2 gas at 400 K
- Cp,m (specific heat capacity at constant pressure) = 2.5 R
- Cv,m (specific heat capacity at constant volume) = 2.5 R
- Volume of vessel A = 5 liters
- Volume of vessel B = 7 liters
- Heat supplied = 5.8 kcal
To find:
- Final pressure in the vessels
Explanation:
1. First, let's calculate the initial pressures in the vessels using the ideal gas law:
- For vessel A: PV = nRT
- P_A * V_A = n_A * R * T_A
- P_A = (n_A * R * T_A) / V_A
- P_A = (2 * R * 300) / 5
- P_A = 120 R/5
- For vessel B: PV = nRT
- P_B * V_B = n_B * R * T_B
- P_B = (n_B * R * T_B) / V_B
- P_B = (4 * R * 400) / 7
- P_B = 1600 R/7
2. Now, let's calculate the change in internal energy for each vessel:
- For vessel A: ΔU_A = n_A * Cv,m * ΔT_A
- ΔU_A = 2 * 2.5R * (T_final - 300)
- For vessel B: ΔU_B = n_B * Cv,m * ΔT_B
- ΔU_B = 4 * 2.5R * (T_final - 400)
3. Since the vessels are connected by a pipe of negligible volume, the total moles of gas remain constant.
- Therefore, n_A + n_B = 2 + 4 = 6 moles
4. The total change in internal energy is the sum of the changes in each vessel:
- ΔU_total = ΔU_A + ΔU_B
- ΔU_total = (2 * 2.5R * (T_final - 300)) + (4 * 2.5R * (T_final - 400))
5. According to the first law of thermodynamics, ΔU_total = q + w, where q is heat supplied and w is work done. Since the vessels are adiabatic (no heat exchange with the surroundings), q = 0.
- Therefore, ΔU_total = w
6. The work done is given by w = -PΔV, where ΔV is the change in volume. The volumes of the vessels remain constant, so ΔV_A = ΔV_B = 0.
7. Therefore, w = 0, and ΔU_total = 0.
8. Plugging in the values, we get:
- (2 * 2.5R * (T_final - 300)) + (4 * 2.5R * (T_final - 400)) = 0
9. Solving the equation, we
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Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.?
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Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.?.
Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.?.
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Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.?, a detailed solution for Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.? has been provided alongside types of Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.? theory, EduRev gives you an
ample number of questions to practice Two rigid adiabatic vessel A and B which initially contain two gases are connected by pipe line with valve of negligible volume.The vessel A contains 2 moles He gas( Cp,m= 2.5 R) at 300 K,vessel B contains 4 moles of O2 gas ( Cv,m= 2.5R) at 400 K .The volume of A and B vessel of A and B is 5 and 7 litre respectively.If the valve is opened and 5.8 kcal heat supplied through it vessels then calculate the final pressure (in atm) in vessels.? tests, examples and also practice JEE tests.
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