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One mole of an ideal monoatomic gas expanded irreversibly in two stage expansion.
State-1 (8.0 bar, 4.0 litre, 300 K)
State-2 (2.0 bar, 16 litre, 300 K)
State-3 (1.0 bar, 32 litre, 300 K)
Total heat absorbed by the gas in the process is :
- a)116 J
- b)40 J
- c)4000 J
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
One mole of an ideal monoatomic gas expanded irreversibly in two stage...
Total work done = – 2(16 – 4) – 1 (32 – 16)
= –24 – 16 = – 40 bar litre
= – 4000 J
q = – w = 4000 J
Most Upvoted Answer
One mole of an ideal monoatomic gas expanded irreversibly in two stage...
Given information:
State 1:
Pressure (P1) = 8.0 bar
Volume (V1) = 4.0 L
Temperature (T1) = 300 K
State 2:
Pressure (P2) = 2.0 bar
Volume (V2) = 16 L
Temperature (T2) = 300 K
State 3:
Pressure (P3) = 1.0 bar
Volume (V3) = 32 L
Temperature (T3) = 300 K
We are given that the gas undergoes an irreversible expansion in two stages. To find the total heat absorbed by the gas in the process, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) supplied to the system minus the work (W) done by the system.
The first law of thermodynamics can be written as:
ΔU = Q - W
In an expansion process, the work done by the gas is given by:
W = -PΔV
where P is the pressure and ΔV is the change in volume.
To find the total heat absorbed, we need to calculate the change in internal energy (ΔU) and the work done (W) for each stage of the expansion.
Stage 1:
In this stage, the gas expands from state 1 (P1, V1) to state 2 (P2, V2) while maintaining a constant temperature (T1 = T2 = 300 K).
ΔU1 = Q1 - W1
Since the process is isothermal, the change in internal energy can be calculated using the equation:
ΔU1 = nCvΔT
where n is the number of moles of the gas and Cv is the molar specific heat capacity at constant volume.
Since the gas is monoatomic, Cv = (3/2)R, where R is the molar gas constant.
ΔT1 = T2 - T1 = 300 K - 300 K = 0 K
ΔU1 = nCvΔT1 = 1 mol * (3/2)R * 0 K = 0 J
The work done in this stage can be calculated as:
W1 = -P1ΔV1
ΔV1 = V2 - V1 = 16 L - 4 L = 12 L
W1 = -P1ΔV1 = -(8.0 bar)(12 L) = -96 L•bar
Since 1 L•bar = 100 J, we can convert the work to joules:
W1 = -96 L•bar * 100 J/L•bar = -9600 J
Therefore, for stage 1:
ΔU1 = 0 J
W1 = -9600 J
Stage 2:
In this stage, the gas expands from state 2 (P2, V2) to state 3 (P3, V3) while maintaining a constant temperature (T2 = T3 = 300 K).
ΔU2 = Q2 - W2
Using the same approach as in stage 1, we can calculate the change in internal energy:
ΔT2 =
State 1:
Pressure (P1) = 8.0 bar
Volume (V1) = 4.0 L
Temperature (T1) = 300 K
State 2:
Pressure (P2) = 2.0 bar
Volume (V2) = 16 L
Temperature (T2) = 300 K
State 3:
Pressure (P3) = 1.0 bar
Volume (V3) = 32 L
Temperature (T3) = 300 K
We are given that the gas undergoes an irreversible expansion in two stages. To find the total heat absorbed by the gas in the process, we can use the first law of thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat (Q) supplied to the system minus the work (W) done by the system.
The first law of thermodynamics can be written as:
ΔU = Q - W
In an expansion process, the work done by the gas is given by:
W = -PΔV
where P is the pressure and ΔV is the change in volume.
To find the total heat absorbed, we need to calculate the change in internal energy (ΔU) and the work done (W) for each stage of the expansion.
Stage 1:
In this stage, the gas expands from state 1 (P1, V1) to state 2 (P2, V2) while maintaining a constant temperature (T1 = T2 = 300 K).
ΔU1 = Q1 - W1
Since the process is isothermal, the change in internal energy can be calculated using the equation:
ΔU1 = nCvΔT
where n is the number of moles of the gas and Cv is the molar specific heat capacity at constant volume.
Since the gas is monoatomic, Cv = (3/2)R, where R is the molar gas constant.
ΔT1 = T2 - T1 = 300 K - 300 K = 0 K
ΔU1 = nCvΔT1 = 1 mol * (3/2)R * 0 K = 0 J
The work done in this stage can be calculated as:
W1 = -P1ΔV1
ΔV1 = V2 - V1 = 16 L - 4 L = 12 L
W1 = -P1ΔV1 = -(8.0 bar)(12 L) = -96 L•bar
Since 1 L•bar = 100 J, we can convert the work to joules:
W1 = -96 L•bar * 100 J/L•bar = -9600 J
Therefore, for stage 1:
ΔU1 = 0 J
W1 = -9600 J
Stage 2:
In this stage, the gas expands from state 2 (P2, V2) to state 3 (P3, V3) while maintaining a constant temperature (T2 = T3 = 300 K).
ΔU2 = Q2 - W2
Using the same approach as in stage 1, we can calculate the change in internal energy:
ΔT2 =
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One mole of an ideal monoatomic gas expanded irreversibly in two stage expansion.State-1 (8.0 bar, 4.0 litre, 300 K)State-2 (2.0 bar, 16 litre, 300 K)State-3 (1.0 bar, 32 litre, 300 K)Total heat absorbed by the gas in the process is :a)116 Jb)40 Jc)4000 Jd)None of theseCorrect answer is option 'C'. Can you explain this answer?
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