NEET Class neet Answered
under isothermal conditions a gas at 300k expand form 0.1l to 0.25l against constant external pressure of 2 bar the work done by the gas is
Asked by deeptideepusingh987 | 23 Oct, 2023, 01:57: PM
Expert Answer
1 L-Bar = 100 J
The temperature of the gas, T= 300 K
Initial volume, V1= 0.1 L
Final volume, V2 = 0.25 L
External pressure, P = 2 bar
Work done by the gas, W = ?
A gas expanding against a constant external pressure is an irreversible process.
The work is done in an irreversible process.
W = -PΔV
W = - P (V2 - V1)
Here,
W is the work done.
P is the pressure.
V is the volume.
The negative (-) sign indicates the system loses energy.
By substituting the given values in the above formula, we get
W = -2 Bar (0.25 L – 0.1 L)
= -2 ⨯ 0.15 L Bar
= -0.30 L Bar (1 L Bar = 100 J)
= - 0.30 ⨯ 100 J
= -30 J
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