Question

If the pressure of an ideal gas contained in a closed vessel is increased by $0.5\%$, the increase in temperature is $2^\circ\text{C}$. The initial temperature of the gas is

• A. $27^\circ\text{C}$
• B. $127^\circ\text{C}$
• C. $300^\circ\text{C}$
• D. $400^\circ\text{C}$

Hint:

When dealing with percentage changes in gas laws, you can also use the differential form: $\frac{\Delta P}{P} = \frac{\Delta T}{T}$. Here, $0.005 = \frac{2}{T}$, leading directly to $T = 400\text{ K}$.

Answer 1

The Correct Option is B

For an ideal gas in a closed vessel, the volume remains constant. Therefore, we apply Gay-Lussac's Law, which states that pressure is directly proportional to absolute temperature. Step 1: Establish the Relationship According to the Ideal Gas Law ($PV = nRT$), at constant volume ($V$): $$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$ Where $T$ must be in Kelvin ($K$).

Step 2: Define Variables Let initial pressure be $P$ and initial temperature be $T$. Final pressure $P_2 = P + 0.005P = 1.005P$. Final temperature $T_2 = T + 2$ (Note: An increase of $2^\circ\text{C}$ is equivalent to an increase of $2\text{ K}$).

Step 3: Solve for $T$ $$\frac{P}{T} = \frac{1.005P}{T + 2}$$ $$1 = \frac{1.005T}{T + 2} \implies T + 2 = 1.005T$$ $$2 = 0.005T$$ $$T = \frac{2}{0.005} = 400\text{ K}\lt strong\gt Step 4: Convert to Celsius\lt /strong\gt t(^\circ\text{C}) = 400 - 273 = 127^\circ\text{C}$$ Thus, the initial temperature was $127^\circ\text{C}$.