A) \[100\text{ }J\]
B) \[150\text{ }J\]
C) \[200\text{ }J\]
D) \[250\text{ }J\]
Correct Answer: A
Solution :
[a] \[dW=25=P({{V}_{2}}-{{V}_{1}})=nR({{T}_{2}}-{{T}_{1}})\] \[dQ=dU+dW-n{{C}_{v}}dT+25\] \[=n.\frac{R}{\gamma -1}({{T}_{2}}-{{T}_{1}})+25\] \[=\frac{25}{\frac{4}{3}-1}+25=100J\left\{ \gamma =\frac{f+2}{f}=\frac{6+2}{6}=\frac{4}{3} \right\}\]
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