At $200 \mathrm{~K}$, an ideal gas $(\mathrm{X})$ present in a $1 \mathrm{~L}$ flask has a concentration of…

At $200 \mathrm{~K}$, an ideal gas $(\mathrm{X})$ present in a $1 \mathrm{~L}$ flask has a concentration of $1 \mathrm{~mol} \mathrm{~L}^{-1}$. At the same temperature, 0.1 mole of $\mathrm{X}$ is added into the vessel: What is the final pressure of the gas in atm? (Given $\mathrm{R}=0.082 \mathrm{~L}^{\text {atom } \mathrm{mol}^{-1} \mathrm{~K}^{-1} \text { ) }}$
  1. 18.04
  2. 16.4
  3. 8.2
  4. 9.02

Solution

At a constant temperature, and volume, $ \begin{aligned} & \frac{\mathrm{P}_1}{\mathrm{n}_1}=\frac{\mathrm{P}_2}{\mathrm{n}_2} \\ & \Rightarrow \mathrm{P}_2=\frac{\mathrm{P}_1}{\mathrm{n}_1} \times \mathrm{n}_2 \end{aligned} $ Now, $\mathrm{P}_1=\frac{\mathrm{n}_1 \mathrm{RT}}{\mathrm{V}}=\frac{(1)(0.082)(200)}{1}=16.4 \mathrm{~atm}$ Thus, $\mathrm{P}_2=\frac{\mathrm{P}_1}{\mathrm{n}_1} \times \mathrm{n}_2=\frac{16.4}{1} \times 1.1$ $ =18.04 \mathrm{~atm} $

Asked in: AP EAMCET 2023 (19 May Shift 1)

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