States of Matter 1 Question 6
6. An open vessel at $27^{\circ} \mathrm{C}$ is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is
(2019 Main, 12 Jan II)
(a) $750 \mathrm{~K}$
(b) $500 \mathrm{~K}$
(c) $750^{\circ} \mathrm{C}$
(d) $500^{\circ} \mathrm{C}$
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Solution:
- Given, temperature $\left(T_{1}\right)=27^{\circ} \mathrm{C}=273+27=300 \mathrm{~K}$
Volume of vessel $=$ constant
Pressure in vessel $=$ constant
Volume of air reduced by $\frac{2}{5}$ so the remaining volume of air is $\frac{3}{5}$.
Let at $T_{1}$ the volume of air inside the vessel is $n$ so at $T_{2}$ the volume of air will be $\frac{3}{5} n$.
Now, as $p$ and $V$ are constant, so
$$ n \cdot T_{1}=\frac{3}{5} n T_{2} $$
Putting the value of $T_{1}$ in equation (i) we get,
$$ \begin{aligned} n \times 300 & =\frac{3}{5} n \times T_{2} \ \text { or } \quad T_{2} & =300 \times \frac{5}{3}=500 \mathrm{~K} \end{aligned} $$
