One mole of a gas at a pressure $2 \mathrm{~Pa}$ and temperature $27^{\circ} \mathrm{C}$ is heated till both…
- $300 \mathrm{~K}$
- $600 \mathrm{~K}$
- $900 \mathrm{~K}$
- $1200 \mathrm{~K}$
Solution
Let volume, $V_1=V$
By using ideal gas equation,
$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}$
Substituting the values, we get
$\frac{2 V}{300}=\frac{4(2 V)}{T} \Rightarrow T=1200 \mathrm{~K}$Asked in: AP EAMCET 2021 (23 Aug Shift 2)