A) 40 K
B) 33 K
C) 20 K
D) 14 K
Correct Answer: C
Solution :
[c] Given: Work done, W= 830 J No. of moles of gas, \[\mu =2\] For diatomic gas \[\gamma =1.4\] Work done during an adiabatic change \[W=\frac{\mu R({{T}_{1}}-{{T}_{2}})}{\gamma -1}\] \[\Rightarrow \] \[830=\frac{2\times 8.3(\Delta T)}{1.4-1}=\frac{2\times 8.3(\Delta T)}{0.4}\] \[\Rightarrow \] \[\Delta T=\frac{830\times 0.4}{2\times 8.3}=20K\]You need to login to perform this action.
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