Question

When the temperature of a gas in a closed vessel is increased by $2.4^\circ\mathrm{C}$, its pressure increases by $0.5%$. The initial temperature of the gas is

• A. $120^\circ\mathrm{C}$
• B. $240^\circ\mathrm{C}$
• C. $480^\circ\mathrm{C}$
• D. $207^\circ\mathrm{C}$

Hint:

Always use Absolute Temperature (Kelvin) in gas law calculations. $T(K) = T(^\circ\mathrm{C}) + 273$.

Answer 1

The Correct Option is D

Step 1: Understand the Gas Law:
For a closed vessel, Volume ($V$) is constant. According to Gay-Lussac's Law, Pressure ($P$) is directly proportional to Absolute Temperature ($T$). \[ P \propto T \implies \frac{\Delta P}{P} = \frac{\Delta T}{T} \] 
Step 2: Substitute Values:
Given: Percentage increase in pressure, $\frac{\Delta P}{P} \times 100 = 0.5 \implies \frac{\Delta P}{P} = 0.005$. Change in temperature $\Delta T = 2.4\,K$ (Change in Celsius is same as Kelvin). Let initial temperature be $T$ Kelvin. \[ 0.005 = \frac{2.4}{T} \] \[ T = \frac{2.4}{0.005} = \frac{2400}{5} = 480\,K \] 
Step 3: Convert to Celsius:
\[ T(^\circ\mathrm{C}) = 480 - 273 = 207^\circ\mathrm{C} \]