A) 1231.5 J
B) 1281.5 J
C) 781.5J
D) 0
Correct Answer: C
Solution :
[c] At constant volume, \[\frac{{{P}_{1}}}{{{T}_{1}}}=\frac{{{P}_{2}}}{{{T}_{2}}}\] \[\Rightarrow {{P}_{2}}=1\times \frac{300}{200}=\frac{3}{2}\] and \[{{V}_{1}}=24.63\text{ }L\] for single phase \[\therefore dG=Vdp-SdT\] \[\Delta G=V\Delta P\int{\left( 2+{{10}^{-2}}T \right)}dt\] \[=1231.5-200-\left( \frac{{{10}^{-2}}\times 50,000}{2} \right)\] \[=781.5\text{ }J\]You need to login to perform this action.
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