A) 540 cal
B) 40cal
C) 500 cal
D) zero
Correct Answer: C
Solution :
Change in volume \[=1671-1=1670c{{m}^{3}}\] \[\therefore \] Work done \[(W)=p.dV\] \[=(1.013\times {{10}^{5}}\times 1670)J\] \[=\frac{1670\times 1.013\times {{10}^{5}}}{4.2}cal\] \[=39.7cal\] Heat given \[(Q)=540cal.\] From first law of thermodynamics, \[\Delta U=Q-W\] \[=540-39.7\] \[=500.3cal\] \[\approx 500cal\]You need to login to perform this action.
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