Air is filled at 60°C in a vessel of open mouth. The vessel is heated to a temperature T so that 1 / 4th part of air escapes. Assuming the volume of vessel remaining constant, the value of T is

The correct option is 171°C.

Step-by-Step Explanation:

Let us understand the behavior of the gas using the ideal gas law:
$PV=nRT$
where:
- $P$ is the pressure of the gas,
- $V$ is the volume of the vessel,
- $n$ is the number of moles of air inside the vessel,
- $R$ is the universal gas constant, and
- $T$ is the absolute temperature in Kelvin (K).

Since the vessel has an open mouth, it is open to the atmosphere, meaning the pressure $P$ remains constant and equal to the atmospheric pressure. The volume of the vessel $V$ is also constant. Since $P$, $V$, and $R$ are constant, we can write:
$nT=\text{constant}$
or,
$n_1{T}_1=n_2{T}_2$

1. Identify the initial state:
Let the initial number of moles of air in the vessel be $n_1$.
The initial temperature is given as $T_1=60°\mathrm{C}$.
Convert this temperature into Kelvin:
$T_1=60+273=333 \mathrm{K}$

2. Identify the final state:
When heated, $\frac{1}{4}\text{th}$ of the air escapes from the vessel. Therefore, the remaining number of moles $n_2$ left in the vessel is:
$n_2=n_1-\frac{1}{4}$

3. Calculate the final temperature $T_2$:
Substitute the values of $T_1$ and $n_2$ into the equation $n_1{T}_1=n_2{T}_2$:
$n_1\times 333=\left(\frac{3}{4}\right)\times T_2$

We can divide both sides by $n_1$:
$333=\frac{3}{4}$

Solving for $T_2$:
$T_2=\frac{333\times 4}{3}$
$T_2=111\times 4=444 \mathrm{K}$

4. Convert the final temperature back to Celsius:
$T(°\mathrm{C})=T_2-273$
$T(°\mathrm{C})=444-273=171°\mathrm{C}$

Thus, the vessel must be heated to a temperature of 171°C.