Question

A gas is heated through \[1^\circ C\] in a closed vessel and so the pressure increases by 0.4%. The initial temperature of the gas was

Answer 1

Hint: Use the equation related to pressure and temperature at constant volume and calculate the initial temperature of the gas. At constant volume pressure of the gas is directly proportional to the temperature of the gas.

Formula Used: \[{\text{P}} \propto {\text{T}}\]
\[\dfrac{{{P_1}}}{{{T_1}}} = \dfrac{{{P_2}}}{{{T_2}}}\]

Complete step by step answer:
We are given gas that is heated through \[1^\circ C\] in a closed vessel so the volume of the system is constant. After the heating pressure of the gas increased by 0.4 %.
Let us assume the initial pressure of the gas as \[{P_1}\] and final pressure of the gas after heating through \[1^\circ C\] as \[{P_2}\].
After heating pressure of the gas increased by 0.4 % so the final pressure of the gas is:
\[{P_2} = {P_1} + 0.04{P_1}\]
Now, we have to assume that the initial temperature of the gas as \[{T_1}\]\[^\circ C\] and the final temperature of the gas as \[{T_2}\].
\[{T_2} = {T_1} + 1^\circ C \]
We must convert this initial and final temperature into Kelvin.
\[{T_1} = {T_1}{\text{ + 273 K }}\]
\[{T_2} = {T_1} +1^\circ C = 273 + {T_1} + {\text{1 = (274 + }}{T_1}{\text{) K}}\]
Now, we can calculate the initial temperature of the gas as follows:
As we know at constant volume pressure of the gas is directly proportional to the temperature of the gas.
So, \[{\text{P}} \propto {\text{T}}\]
\[\dfrac{{{P_1}}}{{{T_1}}} = \dfrac{{{P_2}}}{{{T_2}}}\]
Now, substitute \[{P_1} + 0.04{P_1}\] for \[{P_2}\], \[{\text{(274 + }}{T_1}{\text{) K}}\] for \[{T_2}\] and calculate the initial temperature of the gas.
\[\dfrac{{{P_1}}}{{{T_1}{\text{ + 273 K }}}} = \dfrac{{{P_1} + 0.04{P_1}}}{{{\text{(274 + }}{T_1}{\text{) K}}}}\]
$\Rightarrow$\[\dfrac{{{P_1}}}{{{P_1} + 0.04{P_1}}} = \dfrac{{{T_1}{\text{ + 273 K }}}}{{{\text{(274 + }}{T_1}{\text{) K}}}}\]
$\Rightarrow$\[\dfrac{{{P_1}}}{{{P_1}(1 + 0.04)}} = \dfrac{{{T_1}{\text{ + 273 K }}}}{{{\text{(274 + }}{T_1}{\text{) K}}}}\]
$\Rightarrow$\[\dfrac{1}{{(1 + 0.04)}} = \dfrac{{{T_1}{\text{ + 273 K }}}}{{{\text{(274 + }}{T_1}{\text{) K}}}}\]
$\Rightarrow$\[\dfrac{{{T_1}{\text{ + 273 K }}}}{{{\text{(274 + }}{T_1}{\text{) K}}}} = 0.96\]
$\Rightarrow$\[{T_1}{\text{ + 273 K }} = 0.96{\text{(274 + }}{T_1}{\text{) K}}\]
$\Rightarrow$\[{T_1}{\text{ + 273 K }} = 263{\text{ K + 0}}{\text{.96 }}{T_1}{\text{K}}\]
$\Rightarrow$\[{\text{0}}{\text{.04}} {T_1}{\text{K + 10 K }} = 0\]
\[{T_1}\] = 250 K
Now, we have to convert this initial temperature of the gas from Kelvin to \[^\circ C\].
\[{T_1}\]= (250-273) = -23\[^\circ C\]

So, the initial temperature of the gas was -23.

Note: At constant volume, the pressure of the gas is directly proportional to the temperature of the gas. An increase in temperature increases the pressure of the gas. While the decrease in temperature decreases the pressure of the gas. In calculation always use the values of temperature in Kelvin.