Question

2 mole ideal gas is expanded reversibly from 1 l to 10 l at 227 degree celsius.what is the work done? write answer in calories as well as in joule

Answer 1

Work done in a reversible isothermal process is:

W=−2.303nRTlogViVf.....(1)

Given:-

n=2 moles

T=300K

Vf=10L

Vi=1L

R= Gas constant =8.314J/K−mol

Substituting these values in eqn(1), we have

W=−2.303×2×8.314×300×log110

⇒W=−11488.285J=−11.4kJ

Now as we know that,

ΔH=ΔU+W

For an isothermal process,

ΔU=0

∴ΔH=W=−11.4kJ

Hence the enthalpy change for the given process is −11.4kJ.

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