A) diatomic
B) triatomic
C) a mixture of monoatomic and diatomic
D) monoatomic
Correct Answer: A
Solution :
For adiabatic process, \[dQ=0\] So, \[dU=-\Delta W\] \[\Rightarrow \] \[n{{C}_{V}}dT=+146\times {{10}^{3}}J\] \[\Rightarrow \] \[\frac{nfR}{2}\times 7=146\times {{10}^{3}}\] [\[f\to \]Degree of freedom] \[\Rightarrow \] \[\frac{{{10}^{3}}\times f\times 8.3\times 7}{2}=146\times {{10}^{3}}\] \[f=5.02\approx 5\] So, it is a diatomic gas.You need to login to perform this action.
You will be redirected in
3 sec