Pressure of a gas of constant volume at $20^{\circ} \mathrm{C}$ is $90 \mathrm{~cm}$ of $\mathrm{Hg}$. At…
Pressure of a gas of constant volume at $20^{\circ} \mathrm{C}$ is $90 \mathrm{~cm}$ of $\mathrm{Hg}$. At what temperature the pressure would change to $75 \mathrm{~cm}$ of $\mathrm{Hg}$ ?
$233.2^{\circ} \mathrm{C}$
$-28.8^{\circ} \mathrm{C}$
$-24.2^{\circ} \mathrm{C}$
$28.8^{\circ} \mathrm{C}$
Solution
Given,
$
\begin{aligned}
& T_1=(273+20) \mathrm{K}=293 \mathrm{~K} \\
& p_1=90 \mathrm{~cm} \text { of } \mathrm{Hg} \\
& p_2=75 \mathrm{~cm} \text { of } \mathrm{Hg}
\end{aligned}
$
Since, volume of the gas is constant.
Hence, according to ideal gas equation,
$
\begin{aligned}
\frac{p_1}{T_1} & =\frac{p_2}{T_2} \\
\Rightarrow \quad T_2 & =\frac{T_1 p_2}{p_1}=\frac{293 \times 75}{90} \\
& =244.16 \mathrm{~K}=244.16-273=-28.8^{\circ} \mathrm{C}
\end{aligned}
$