A) diatomic
B) triatomic
C) a mixture of monatomic and diatomic
D) monatomic
Correct Answer: A
Solution :
For adiabatic process, \[\Delta Q=0\] So, \[\Delta U=-\Delta W\] (from 1st law,\[\Delta Q=\Delta U+\Delta W\]) \[\Rightarrow \]\[n{{C}_{v}}dT=+146\times {{10}^{3}}J\] \[\Rightarrow \]\[\frac{nfR}{2}\times 7=146\times {{10}^{3}}\] (\[\because \]\[{{C}_{v}}=\frac{1}{2}nfR,\]where,\[f\]is 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.
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