A) 4p0V0
B) 6p0V0
C) 4.5p0V0
D) 2p0V0
Correct Answer: B
Solution :
Change in internal energy from A to B is \[\Delta U=\frac{f}{2}\mu R\Delta T=\frac{f}{2}({{p}_{f}}{{V}_{f}}-{{p}_{i}}{{V}_{i}})\] \[=\frac{3}{2}(2{{p}_{0}}\times 2{{V}_{0}}-{{p}_{0}}\times {{V}_{0}})\] \[=\frac{9}{2}{{p}_{0}}{{V}_{0}}\] Work done in process A to B is equal to the area covered by the graph with volume axis, ie, \[{{W}_{A\to B}}=\frac{1}{2}({{p}_{0}}+2{{p}_{0}})\times (2{{V}_{0}}-{{V}_{0}})=\frac{3}{2}{{p}_{0}}{{V}_{0}}\] Hence, \[\Delta Q=\Delta U+\Delta W\] \[=\frac{9}{2}{{p}_{0}}{{V}_{0}}+\frac{3}{2}{{p}_{0}}{{V}_{0}}\] \[=6{{p}_{0}}{{V}_{0}}\]
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