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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 $\mathrm{R}=0.082 \mathrm{~L}^{\text {atom } \mathrm{mol}^{-1} \mathrm{~K}^{-1} \text { ) }}$
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(Given $\mathrm{R}=0.082 \mathrm{~L}^{\text {atom } \mathrm{mol}^{-1} \mathrm{~K}^{-1} \text { ) }}$
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1709 Upvotes
Verified Answer
The correct answer is:
18.04
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, $\mathrm{P}_2=\frac{\mathrm{P}_1}{\mathrm{n}_1} \times \mathrm{n}_2=\frac{16.4}{1} \times 1.1$
$$
=18.04 \mathrm{~atm}
$$
$$
\begin{aligned}
& \frac{\mathrm{P}_1}{\mathrm{n}_1}=\frac{\mathrm{P}_2}{\mathrm{n}_2} \\
& \Rightarrow \mathrm{P}_2=\frac{\mathrm{P}_1}{\mathrm{n}_1} \times \mathrm{n}_2
\end{aligned}
$$
Now, $\mathrm{P}_1=\frac{\mathrm{n}_1 \mathrm{RT}}{\mathrm{V}}=\frac{(1)(0.082)(200)}{1}=16.4 \mathrm{~atm}$
Thus, $\mathrm{P}_2=\frac{\mathrm{P}_1}{\mathrm{n}_1} \times \mathrm{n}_2=\frac{16.4}{1} \times 1.1$
$$
=18.04 \mathrm{~atm}
$$
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