Question

Argon gas taken in a closed vessel was heated from $ - 13^\circ C$ to $13^\circ C$, the percentage increase in pressure inside the bulb is
(A) 11%
(B) 10%
(C) 100%
(D) 110%

Answer 1

Hint: To determine the percentage increase in pressure, we will use the ideal gas equation to get information about temperature and pressure variation. Also, convert the values of the temperature into Kelvin and then put it into the obtained relation.

Complete step by step answer:
The argon gas's final and initial temperature is given in the question, so we can use the ideal gas equation to determine the increase in pressure.
From the ideal gas equation, write the relation between temperature and pressure.
$p \propto T$
Here is the pressure of the gas, and $T$ is the temperature of the gas.
In the above equation, it is clear that temperature is directly proportional to the pressure; therefore, we get
$\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$...... (1)
Here, ${P_1}$ is the initial pressure of the argon gas, ${P_2}$ is the final temperature of the argon gas, ${T_1}$ is the initial temperature of the argon gas, and ${T_2}$ is the final temperature of the argon gas.
From equation (1), we will write the expression of the percentage increase in pressure of argon gas, so
\[\dfrac{{{P_2} - {p_1}}}{{{p_1}}} \times 100\% = \dfrac{{{T_2} - {T_1}}}{{{T_1}}} \times 100\% \]...... (2)
To convert the temperature into Kelvin, we will add 273 into the values of the temperature, so we get
$\begin{array}{l}
{T_1} = - 13^\circ C + 273\;K\\
{T_1} = 260\;K
\end{array}$
And
$\begin{array}{l}
{T_2} = 13^\circ C + 273\,K\\
{T_2} = 286\;K
\end{array}$
Now use these values in equation (2) to calculate the percentage increase in argon gas pressure.
Therefore, we get
\[\begin{array}{l}
\dfrac{{{P_2} - {p_1}}}{{{p_1}}} \times 100\% = \dfrac{{286\;K - 260\,K}}{{260\,K}} \times 100\% \\
\Rightarrow \dfrac{{{P_2} - {p_1}}}{{{p_1}}} \times 100\% = \dfrac{{26\,K}}{{260\,K}} \times 100\% \\
\Rightarrow \dfrac{{{P_2} - {p_1}}}{{{p_1}}} \times 100\% = 0.1 \times 100\% \\
\Rightarrow \dfrac{{{P_2} - {p_1}}}{{{p_1}}} \times 100\% = 10\%
\end{array}\]
Therefore, the percentage increase in pressure inside the bulb is $10\% $, and option (B) is correct.

Note: we can also determine the correct answer from the information that the gas's temperature relates directly to the gas's pressure for the closed volume. It means that if any quantity increases, the other one will decrease and vice versa.