Answer:
Option D
Explanation:
given, initial temperature of 1 mole of $N_{2}$ gas
$T_{0}$ =300K
initial pressure of gas , $p_{1}$=p
and final pressure of gas , $p_{2}=10p$
By adiabatic process relation between p and T.
$p_{1}^{1-\gamma}T_{1}^{\gamma}=p_{2}^{1-\gamma}T_{2}^{\gamma}$
$\left(\frac{p_{1}}{p_{2}}\right)^{1-\gamma}=\left(\frac{T_{2}}{T_{1}}\right)^{\gamma}\Rightarrow\left(\frac{p}{10p}\right)^{1-\gamma}=\left(\frac{T_{2}}{300}\right)^{\gamma}$
$10^{\gamma-1}=\frac{T_{2}^{\gamma}}{300^{\gamma}}\Rightarrow T_{2}^{\gamma}= 10^{\gamma-1}\times 300^{\gamma}$
$\Rightarrow$ $T_{2}=10^{\frac{\gamma-1}{\gamma}}\times 300= 10^{1-\frac{1}{\gamma}}\times300$
$= 10^{1-\frac{1}{7/5}}\times300\left[\because For diatomic,\gamma=\frac{7}{5}\right]$
$= 10^{2/7}\times300= (100)^{1/7} \times 300$
=$1.9 \times 300$ [$\because 100^{3/7}=1.9$]
=570 K
Therefore, the final gas temperature after compression is 570 K
