Question

The work of 146kJ is performed in order to compress one kilo mole of a gas adiabatically and in this process the temperature of the gas increases by 7°. The gas is
1)diatomic 2)triatomic 3)monoatomic 4)mixture of monoatomic and diatomic
Ans-(1)

Answer 1

Dear student

The work done in adiabatic process is given by
$$\begin{gathered} W=\frac{\mu R\left(T_1-T_2\right)}{\left(\gamma -1\right)} \\ where \\ \mu ={10}^{3}mole \\ {T}_1-T_2=7^{o}C=7K \\ R=8.31J/moleK \\ so \\ 146\times {10}^3=\frac{{10}^3\times 8.31\times 7}{\left(\gamma -1\right)} \\ \left(\gamma -1\right)=\frac{{10}^3\times 8.31\times 7}{146\times {10}^3} \\ \gamma -1\approx 0.4 \\ but\gamma =1+\frac{2}{f}where\gamma istheratiooftwospecificheatsofagas \\ andfisthedegreeoffreedomofthegas \\ 1+\frac{2}{f}-1=0.4 \\ \frac{2}{f}=0.4 \\ f=5 \\ sothegaswillbediatomicbecausedegreeoffreedomfordiatomic=5 \end{gathered}$$