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Question
Two moles of a monoatomic gas at 300 K is kept in a non-conducting container enclosed by a piston. Gas is now compressed to increase the temperature from 300 K to 400 K. Find work done by the gas (R=253J/mol K)
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Solution
ΔQ=ΔU+W
Since container is non conducting therefore
ΔQ=0=ΔU+W
⇒W=−ΔU=−nf2RΔT=−2×32R(400−300)
=−3×253×100J=−2500J
Since container is non conducting therefore
ΔQ=0=ΔU+W
⇒W=−ΔU=−nf2RΔT=−2×32R(400−300)
=−3×253×100J=−2500J
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