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1 mole of an ideal gas A(Cv, m = 3R) and 2 moles of an ideal gas B are (Cv,m = 3/2 R) taken in a container and expanded reversible and adiabatically from 1 liter to 4 liters starting from initial temperature of 320 K. ΔE or ΔU For the process is :
- a)−140R
- b)−240R
- c)−480R
- d)−960R
Correct answer is option 'D'. Can you explain this answer?
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1 mole of an ideal gas A(Cv, m = 3R) and 2 moles of an ideal gas B ar...
ΔE or ΔU for the process can be calculated using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat transferred to the system minus the work done by the system.
The process is reversible and adiabatic, which means that there is no heat transfer (Q = 0) and the work done by the system is equal to the change in internal energy (W = ΔU).
To calculate the change in internal energy (ΔU), we can use the formula:
ΔU = nCvΔT
where n is the number of moles of the gas and Cv is the molar heat capacity at constant volume.
Since we have 1 mole of gas A and 2 moles of gas B, the total number of moles (n) is 1 + 2 = 3.
For gas A, Cv,m = 3R, so Cv = 3R/1 mol = 3R.
For gas B, Cv,m = 3/2 R, so Cv = (3/2)R/1 mol = (3/2)R.
The initial temperature (T1) is 320 K and the final temperature (T2) can be calculated using the ideal gas law:
P1V1/T1 = P2V2/T2
Since the process is adiabatic, the pressure (P) and volume (V) are inversely proportional:
P1V1^(γ) = P2V2^(γ)
where γ is the heat capacity ratio, which is equal to Cv,m/R for an ideal gas.
For gas A, γ = Cv,m/R = 3R/R = 3.
For gas B, γ = Cv,m/R = (3/2)R/R = 3/2.
Solving for P2/P1, we get:
P2/P1 = (V1/V2)^(γ) = (1/4)^(3/2) = 1/8
Using the ideal gas law, we can calculate the final temperature (T2):
P2V2/T2 = P1V1/T1
(1/8)(4 L)/T2 = (1 atm)(1 L)/320 K
T2 = (1/8)(4 L)(320 K)/(1 L) = 40 K
Now we can calculate the change in internal energy (ΔU) for each gas using the formula mentioned earlier:
ΔU_A = n_A * Cv_A * ΔT_A = (1 mol)(3R)(40 K - 320 K) = -960R
ΔU_B = n_B * Cv_B * ΔT_B = (2 mol)(3/2 R)(40 K - 320 K) = -960R
The total change in internal energy (ΔU) is the sum of the changes in internal energy for each gas:
ΔU = ΔU_A + ΔU_B = -960R + -960R = -1920R
Since the question asks for ΔE or ΔU, the answer is -1920R, which is closest to option 'D' (-960R).
The process is reversible and adiabatic, which means that there is no heat transfer (Q = 0) and the work done by the system is equal to the change in internal energy (W = ΔU).
To calculate the change in internal energy (ΔU), we can use the formula:
ΔU = nCvΔT
where n is the number of moles of the gas and Cv is the molar heat capacity at constant volume.
Since we have 1 mole of gas A and 2 moles of gas B, the total number of moles (n) is 1 + 2 = 3.
For gas A, Cv,m = 3R, so Cv = 3R/1 mol = 3R.
For gas B, Cv,m = 3/2 R, so Cv = (3/2)R/1 mol = (3/2)R.
The initial temperature (T1) is 320 K and the final temperature (T2) can be calculated using the ideal gas law:
P1V1/T1 = P2V2/T2
Since the process is adiabatic, the pressure (P) and volume (V) are inversely proportional:
P1V1^(γ) = P2V2^(γ)
where γ is the heat capacity ratio, which is equal to Cv,m/R for an ideal gas.
For gas A, γ = Cv,m/R = 3R/R = 3.
For gas B, γ = Cv,m/R = (3/2)R/R = 3/2.
Solving for P2/P1, we get:
P2/P1 = (V1/V2)^(γ) = (1/4)^(3/2) = 1/8
Using the ideal gas law, we can calculate the final temperature (T2):
P2V2/T2 = P1V1/T1
(1/8)(4 L)/T2 = (1 atm)(1 L)/320 K
T2 = (1/8)(4 L)(320 K)/(1 L) = 40 K
Now we can calculate the change in internal energy (ΔU) for each gas using the formula mentioned earlier:
ΔU_A = n_A * Cv_A * ΔT_A = (1 mol)(3R)(40 K - 320 K) = -960R
ΔU_B = n_B * Cv_B * ΔT_B = (2 mol)(3/2 R)(40 K - 320 K) = -960R
The total change in internal energy (ΔU) is the sum of the changes in internal energy for each gas:
ΔU = ΔU_A + ΔU_B = -960R + -960R = -1920R
Since the question asks for ΔE or ΔU, the answer is -1920R, which is closest to option 'D' (-960R).
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1 mole of an ideal gas A(Cv, m = 3R) and 2 moles of an ideal gas B ar...
for adiabatic process ΔU = ΔW
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1 mole of an ideal gas A(Cv, m = 3R) and 2 moles of an ideal gas B are (Cv,m = 3/2 R) taken in a container and expanded reversible and adiabatically from 1 liter to 4 liters starting from initial temperature of 320 K. ΔE or ΔU For the process is :a)−140Rb)−240Rc)−480Rd)−960RCorrect answer is option 'D'. Can you explain this answer?
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1 mole of an ideal gas A(Cv, m = 3R) and 2 moles of an ideal gas B are (Cv,m = 3/2 R) taken in a container and expanded reversible and adiabatically from 1 liter to 4 liters starting from initial temperature of 320 K. ΔE or ΔU For the process is :a)−140Rb)−240Rc)−480Rd)−960RCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1 mole of an ideal gas A(Cv, m = 3R) and 2 moles of an ideal gas B are (Cv,m = 3/2 R) taken in a container and expanded reversible and adiabatically from 1 liter to 4 liters starting from initial temperature of 320 K. ΔE or ΔU For the process is :a)−140Rb)−240Rc)−480Rd)−960RCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of an ideal gas A(Cv, m = 3R) and 2 moles of an ideal gas B are (Cv,m = 3/2 R) taken in a container and expanded reversible and adiabatically from 1 liter to 4 liters starting from initial temperature of 320 K. ΔE or ΔU For the process is :a)−140Rb)−240Rc)−480Rd)−960RCorrect answer is option 'D'. Can you explain this answer?.
1 mole of an ideal gas A(Cv, m = 3R) and 2 moles of an ideal gas B are (Cv,m = 3/2 R) taken in a container and expanded reversible and adiabatically from 1 liter to 4 liters starting from initial temperature of 320 K. ΔE or ΔU For the process is :a)−140Rb)−240Rc)−480Rd)−960RCorrect answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about 1 mole of an ideal gas A(Cv, m = 3R) and 2 moles of an ideal gas B are (Cv,m = 3/2 R) taken in a container and expanded reversible and adiabatically from 1 liter to 4 liters starting from initial temperature of 320 K. ΔE or ΔU For the process is :a)−140Rb)−240Rc)−480Rd)−960RCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 1 mole of an ideal gas A(Cv, m = 3R) and 2 moles of an ideal gas B are (Cv,m = 3/2 R) taken in a container and expanded reversible and adiabatically from 1 liter to 4 liters starting from initial temperature of 320 K. ΔE or ΔU For the process is :a)−140Rb)−240Rc)−480Rd)−960RCorrect answer is option 'D'. Can you explain this answer?.
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