Q21 / 180 · COMEDK 2023 (Evening Shift) · Chemistry
A Gas taken in a closed vessel is heated from $54^{\circ} \mathrm{C}$ to $1254^{\circ} \mathrm{C}$. The pressure of the gas becomes ___________ times its original pressure $\mathrm{P_1}$
- A23.2 times $\mathrm{P_1}$
- B9.6 times $\mathrm{P_1}$
- Correct answerC4.67 times $\mathrm{P_1}$
- D8.92 times $\mathrm{P_1}$
Solution
For a gas in a closed vessel, the volume $V$ remains constant. According to Gay-Lussac's Law, for a fixed amount of gas at constant volume, the pressure $P$ is directly proportional to the absolute temperature $T$.
The initial temperature $T_1 = 54^{\circ} \mathrm{C} = 54 + 273 = 327 \mathrm{K}$.
The final temperature $T_2 = 1254^{\circ} \mathrm{C} = 1254 + 273 = 1527 \mathrm{K}$.
Using the relation $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$, we have $P_2 = P_1 \times \dfrac{T_2}{T_1}$.
$P_2 = P_1 \times \dfrac{1527}{327}$.
Calculating the ratio: $\dfrac{1527}{327} \approx 4.67$.
Therefore, the pressure becomes $4.67$ times its original pressure $P_1$.
Answer: 4.67 times $\mathrm{P_1}$
The initial temperature $T_1 = 54^{\circ} \mathrm{C} = 54 + 273 = 327 \mathrm{K}$.
The final temperature $T_2 = 1254^{\circ} \mathrm{C} = 1254 + 273 = 1527 \mathrm{K}$.
Using the relation $\dfrac{P_1}{T_1} = \dfrac{P_2}{T_2}$, we have $P_2 = P_1 \times \dfrac{T_2}{T_1}$.
$P_2 = P_1 \times \dfrac{1527}{327}$.
Calculating the ratio: $\dfrac{1527}{327} \approx 4.67$.
Therefore, the pressure becomes $4.67$ times its original pressure $P_1$.
Answer: 4.67 times $\mathrm{P_1}$
Hence, the correct answer is option: 'C'.