Two liters of N, at 0°C and 5 atm pressure is expanded isothermally against a constant external pressure of 1 atm until the pressure of gas reaches 1 atm. Assuming gas to be ideal, calculate work of expansion.

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answer is 810.6.

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Detailed Solution

Since the external pressure is greatly different from the pressure of N₂ and thus, process is irreversible. By Boyle's law,

P₁V₁=P₂V₂

P₁=5atm, P₂=1atm, V1=2L

V₂= (5atm×2L)/1atm

V₂=10L

Now, Work -w=Pext.ΔV

w=−1atm (10−2) L (Ext. pressure =1atm)

w=−8atm.L

Therefore,

−8atm.L=(1atmL101.325) ×8=−810.6J (1atmL=101.325)

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