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An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.
Imp : you should answer only the numeric value
Correct answer is '373'. Can you explain this answer?
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An ideal gas is adiabatically and irreversibly compressed from 3 bar a...
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Given:
Initial pressure (P1) = 3 bar
Initial temperature (T1) = 300 K
Final pressure (P2) = 6 bar
Molar heat capacity at constant volume (Cv) = 30 J mol-1 K-1
Ratio of molar heat capacities (γ) = 1.277
Universal gas constant (R) = 8.314 J mol-1 K-1
Work done irreversibly (W_irrev) = 1.5 * Work done reversibly (W_rev)
To find: Final temperature (T2) in the irreversible compression case
Formula used:
γ = Cp/Cv
Cp - Cv = R
First law of thermodynamics: ΔE = Q - W
Solution:
1. Finding the work done in the reversible compression case:
In reversible compression, the process is isentropic (adiabatic) and the gas follows the equation PV^γ = constant.
Using this equation, we can find the final volume (V2) of the gas:
(P1V1^γ) = (P2V2^γ)
V2 = V1 * (P1/P2)^(1/γ)
Using the formula for work done in an isentropic process:
W_rev = Cv * (T1 - T2) * (1 - (P2/P1)^((γ-1)/γ))
Substituting the given values:
W_rev = 30 * (300 - T2) * (1 - (6/3)^(0.277))
W_rev = 30 * (300 - T2) * 0.093
2. Finding the work done in the irreversible compression case:
From the given data, we know that W_irrev = 1.5 * W_rev
3. Finding the change in internal energy (ΔE):
Since the process is adiabatic (Q = 0), ΔE = -W
4. Finding the final temperature (T2) in the irreversible compression case:
Using the first law of thermodynamics:
ΔE_irrev = -W_irrev = -1.5 * W_rev
ΔE_rev = -W_rev
ΔE_irrev = ΔE_rev
Cv * (T1 - T2_irrev) = Cv * (T1 - T2_rev)
T2_irrev = T1 - (T1 - T2_rev)/1.5
Substituting the values of T2_rev and W_rev in the above equation:
T2_irrev = 300 - (300 - T2) * 0.667/0.093
Simplifying:
T2_irrev = 373.1 K
Rounding off to one decimal place:
T2_irrev = 373 K
Therefore, the final temperature of the gas in the irreversible compression case is 373 K.
Initial pressure (P1) = 3 bar
Initial temperature (T1) = 300 K
Final pressure (P2) = 6 bar
Molar heat capacity at constant volume (Cv) = 30 J mol-1 K-1
Ratio of molar heat capacities (γ) = 1.277
Universal gas constant (R) = 8.314 J mol-1 K-1
Work done irreversibly (W_irrev) = 1.5 * Work done reversibly (W_rev)
To find: Final temperature (T2) in the irreversible compression case
Formula used:
γ = Cp/Cv
Cp - Cv = R
First law of thermodynamics: ΔE = Q - W
Solution:
1. Finding the work done in the reversible compression case:
In reversible compression, the process is isentropic (adiabatic) and the gas follows the equation PV^γ = constant.
Using this equation, we can find the final volume (V2) of the gas:
(P1V1^γ) = (P2V2^γ)
V2 = V1 * (P1/P2)^(1/γ)
Using the formula for work done in an isentropic process:
W_rev = Cv * (T1 - T2) * (1 - (P2/P1)^((γ-1)/γ))
Substituting the given values:
W_rev = 30 * (300 - T2) * (1 - (6/3)^(0.277))
W_rev = 30 * (300 - T2) * 0.093
2. Finding the work done in the irreversible compression case:
From the given data, we know that W_irrev = 1.5 * W_rev
3. Finding the change in internal energy (ΔE):
Since the process is adiabatic (Q = 0), ΔE = -W
4. Finding the final temperature (T2) in the irreversible compression case:
Using the first law of thermodynamics:
ΔE_irrev = -W_irrev = -1.5 * W_rev
ΔE_rev = -W_rev
ΔE_irrev = ΔE_rev
Cv * (T1 - T2_irrev) = Cv * (T1 - T2_rev)
T2_irrev = T1 - (T1 - T2_rev)/1.5
Substituting the values of T2_rev and W_rev in the above equation:
T2_irrev = 300 - (300 - T2) * 0.667/0.093
Simplifying:
T2_irrev = 373.1 K
Rounding off to one decimal place:
T2_irrev = 373 K
Therefore, the final temperature of the gas in the irreversible compression case is 373 K.
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An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.Imp : you should answer only the numeric valueCorrect answer is '373'. Can you explain this answer?
Question Description
An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.Imp : you should answer only the numeric valueCorrect answer is '373'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.Imp : you should answer only the numeric valueCorrect answer is '373'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.Imp : you should answer only the numeric valueCorrect answer is '373'. Can you explain this answer?.
An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.Imp : you should answer only the numeric valueCorrect answer is '373'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.Imp : you should answer only the numeric valueCorrect answer is '373'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.Imp : you should answer only the numeric valueCorrect answer is '373'. Can you explain this answer?.
Solutions for An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.Imp : you should answer only the numeric valueCorrect answer is '373'. Can you explain this answer? in English & in Hindi are available as part of our courses for GATE.
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An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.Imp : you should answer only the numeric valueCorrect answer is '373'. Can you explain this answer?, a detailed solution for An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.Imp : you should answer only the numeric valueCorrect answer is '373'. Can you explain this answer? has been provided alongside types of An ideal gas is adiabatically and irreversibly compressed from 3 bar and 300 K to 6 bar in a closed system. The work required for the irreversible compression is 1.5 times the work that is required for reversible compression from the same initial temperature and pressure to the same final pressure. The molar heat capacity of the gas at constant volume is 30 J mol-1 K-1 (assumed to be independent of temperature); universal gas constant, R is 8.314 J mol-1 K-1; ratio of molar heat capacities is 1.277. The temperature (in K, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is __________.Imp : you should answer only the numeric valueCorrect answer is '373'. Can you explain this answer? theory, EduRev gives you an
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