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A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.?
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A rigid insulated tank of volume 4m3 is divided into two compartments ...
Problem Statement:
A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.
Solution:
To calculate the total change in entropy of the process, we can use the equation:
ΔS = ΔS_A + ΔS_B
where ΔS_A is the change in entropy of gas A and ΔS_B is the change in entropy of gas B.
Change in entropy of gas A (ΔS_A):
The change in entropy of gas A can be calculated using the equation:
ΔS_A = C_vA * ln(TfA / TiA) + R * ln(VfA / ViA)
where C_vA is the specific heat capacity at constant volume of gas A, TfA is the final temperature of gas A, TiA is the initial temperature of gas A, VfA is the final volume of gas A, and ViA is the initial volume of gas A.
Given:
C_vA = 1.4 (specific heat capacity ratio)
TiA = 400 K
TfA = (Total energy of the system) / (Total number of moles of gas A) = [(C_vA * TiA) / (C_vA + R)] = [(1.4 * 400) / (1.4 + 0.287)] = 632.96 K
ViA = 2 m^3
VfA = 4 m^3
Substituting these values into the equation, we get:
ΔS_A = 1.4 * ln(632.96 / 400) + 0.287 * ln(4 / 2)
Calculating this, we find that ΔS_A ≈ 0.605 kJ/K
Change in entropy of gas B (ΔS_B):
Similarly, the change in entropy of gas B can be calculated using the same equation:
ΔS_B = C_vB * ln(TfB / TiB) + R * ln(VfB / ViB)
Given:
C_vB = 1.4 (specific heat capacity ratio)
TiB = 700 K
TfB = (Total energy of the system) / (Total number of moles of gas B) = [(C_vB * TiB) / (C_vB + R)] = [(1.4 * 700) / (1.4 + 0.287)] = 1104.05 K
ViB = 2 m^3
VfB = 4 m^3
Substituting these values into the equation, we get:
ΔS_B = 1.4 * ln(1104.05 / 700) + 0.287 * ln(4 / 2)
Calculating this, we find that ΔS_B ≈ 1.041 kJ/K
Total change in entropy (ΔS):
A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.
Solution:
To calculate the total change in entropy of the process, we can use the equation:
ΔS = ΔS_A + ΔS_B
where ΔS_A is the change in entropy of gas A and ΔS_B is the change in entropy of gas B.
Change in entropy of gas A (ΔS_A):
The change in entropy of gas A can be calculated using the equation:
ΔS_A = C_vA * ln(TfA / TiA) + R * ln(VfA / ViA)
where C_vA is the specific heat capacity at constant volume of gas A, TfA is the final temperature of gas A, TiA is the initial temperature of gas A, VfA is the final volume of gas A, and ViA is the initial volume of gas A.
Given:
C_vA = 1.4 (specific heat capacity ratio)
TiA = 400 K
TfA = (Total energy of the system) / (Total number of moles of gas A) = [(C_vA * TiA) / (C_vA + R)] = [(1.4 * 400) / (1.4 + 0.287)] = 632.96 K
ViA = 2 m^3
VfA = 4 m^3
Substituting these values into the equation, we get:
ΔS_A = 1.4 * ln(632.96 / 400) + 0.287 * ln(4 / 2)
Calculating this, we find that ΔS_A ≈ 0.605 kJ/K
Change in entropy of gas B (ΔS_B):
Similarly, the change in entropy of gas B can be calculated using the same equation:
ΔS_B = C_vB * ln(TfB / TiB) + R * ln(VfB / ViB)
Given:
C_vB = 1.4 (specific heat capacity ratio)
TiB = 700 K
TfB = (Total energy of the system) / (Total number of moles of gas B) = [(C_vB * TiB) / (C_vB + R)] = [(1.4 * 700) / (1.4 + 0.287)] = 1104.05 K
ViB = 2 m^3
VfB = 4 m^3
Substituting these values into the equation, we get:
ΔS_B = 1.4 * ln(1104.05 / 700) + 0.287 * ln(4 / 2)
Calculating this, we find that ΔS_B ≈ 1.041 kJ/K
Total change in entropy (ΔS):
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A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.?
Question Description
A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.?.
A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.?.
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A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.?, a detailed solution for A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.? has been provided alongside types of A rigid insulated tank of volume 4m3 is divided into two compartments by a removable partition. One compartment of volume 2m3 contains ideal gas A at 400 K and 5bar while the other compartment contains ideal gas B at 700 K and 15 bar. The partition is removed and the gas is allowed to mix. After mixing, calculate the total change in entropy of the process. Given, specific heat capacity ratio of both ideal gases is 1.4.? theory, EduRev gives you an
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