3.4 moles of an ideal gas occupies volume of 68 mL at 300 K . What would be the pressure of gas?…
- $1.247 \times 10^2 \mathrm{kPa}$
- $2.431 \times 10^3 \mathrm{kPa}$
- $1.031 \times 10^5 \mathrm{kPa}$
- $3.247 \times 10^5 \mathrm{kPa}$
Solution
According to ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $\begin{aligned} \therefore \quad \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{~V}} & =\frac{3.4 \mathrm{~mol} \times 8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1} \times 300 \mathrm{~K}}{0.068 \mathrm{dm}^3} \\ & =124710 \mathrm{kPa} \end{aligned}$
Asked in: MHT CET 2024 (15 May Shift 2)