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A vessel is filled with an ideal gas at a pressure of 10 atm and temperature $27^{\circ} \mathrm{C} .$ Half of the mass is removed from the vessel and temperature of the remaining gas is increased to $87^{\circ} \mathrm{C}$. Then the pressure of the gas in the vessel will be
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Verified Answer
The correct answer is:
$6 \mathrm{~atm}$
Ideal gas equation
$$
\begin{array}{c}
p V=n R T \\
\frac{p V}{T}=n R=\frac{m}{M} R \\
\frac{m}{M} R=\frac{10 \times V}{300} \\
\frac{m}{2 M} R=\frac{p V}{360}
\end{array}
$$
$$
\begin{aligned}
\frac{p V}{360} &=\frac{1}{2} \frac{10 \mathrm{~V}}{300} \\
p &=\frac{3600}{2 \times 300}=6 \mathrm{~atm}
\end{aligned}
$$
$$
\begin{array}{c}
p V=n R T \\
\frac{p V}{T}=n R=\frac{m}{M} R \\
\frac{m}{M} R=\frac{10 \times V}{300} \\
\frac{m}{2 M} R=\frac{p V}{360}
\end{array}
$$
$$
\begin{aligned}
\frac{p V}{360} &=\frac{1}{2} \frac{10 \mathrm{~V}}{300} \\
p &=\frac{3600}{2 \times 300}=6 \mathrm{~atm}
\end{aligned}
$$
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