In an isochoric process if $t_1=27^{\circ} \mathrm{C}$ and, $t_2=127^{\circ} \mathrm{C}$ then…
In an isochoric process if $t_1=27^{\circ} \mathrm{C}$ and, $t_2=127^{\circ} \mathrm{C}$ then $\frac{P_1}{P_2}$ will be equal to $\left[P_1\right.$ and $P_2$ are the pressures at $t_1{ }^{\circ} \mathrm{C}$ and $t_2^{\circ} \mathrm{C}$ respectively]
$\frac{9}{59}$
$\frac{4}{3}$
$\frac{3}{4}$
$\frac{2}{3}$
Solution
At constant volume $P \propto T \therefore \frac{P_1}{P_2}=\frac{T_1}{T_2}=\frac{300}{400}=\frac{3}{4}$