Workdonebythegasduringexpansionatconstantpressure
∆W=−pdV
∆W=−pd(V2−V1)
∆W=−1.60×105pa(0.320m3−0.120m3)
=−0.32×105pa.m3
=−32000pa.m3
=−32000(N/m2).(m3)
=−32000N.m
∆W=−32000Joule
Workdonebythesystem=−32000Joule
(b)Changeintheinternalenergyofthegas
∆U=∆Q+∆W
Given
∆Q=1.15×105Joule
∆W=−32000Joule
∆U=1.15×105J−32000Joule
∆U=83000JouleAns.
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