Q43 / 150 · MHT CET 2026 (17 April Shift 1) · Chemistry
$V_0$ and $V_t$ are volumes of an ideal gas at $0\ ^\circ\text{C}$ and $t\ ^\circ\text{C}$ respectively, find ratio $\dfrac{V_t}{V_0}$
- A$\dfrac{1}{273} \times t$
- B$1 + 273 \times t$
- Correct answerC$1 + \dfrac{t}{273}$
- D$\dfrac{1}{273 + t}$
Solution
According to Charles's law, at constant pressure, the volume of a fixed mass of an ideal gas is directly proportional to its absolute temperature.
$V \propto T$
$\dfrac{V_t}{V_0} = \dfrac{T_t}{T_0}$
The absolute temperatures are $T_0 = 273\text{ K}$ and $T_t = (273 + t)\text{ K}$.
$\dfrac{V_t}{V_0} = \dfrac{273 + t}{273}$
$\dfrac{V_t}{V_0} = 1 + \dfrac{t}{273}$
Answer: $1 + \dfrac{t}{273}$
$V \propto T$
$\dfrac{V_t}{V_0} = \dfrac{T_t}{T_0}$
The absolute temperatures are $T_0 = 273\text{ K}$ and $T_t = (273 + t)\text{ K}$.
$\dfrac{V_t}{V_0} = \dfrac{273 + t}{273}$
$\dfrac{V_t}{V_0} = 1 + \dfrac{t}{273}$
Answer: $1 + \dfrac{t}{273}$
Hence, the correct answer is option: 'C'.