Q115 / 150 · MHT CET 2026 (19 April Shift 1) · Physics
The kinetic energy of an ideal gas is $E_0$ at $27^\circ\text{C}$. When the temperature is increased to $177^\circ\text{C}$, then kinetic energy will be
- A$E_0 / 2$
- B$2E_0 / 3$
- Correct answerC$3E_0 / 2$
- D$2E_0$
Solution
The kinetic energy of an ideal gas is directly proportional to its absolute temperature.
$E \propto T$
Initial temperature $T_1 = 27^\circ\text{C} = 27 + 273 = 300\text{ K}$
Final temperature $T_2 = 177^\circ\text{C} = 177 + 273 = 450\text{ K}$
Using the relation $\dfrac{E_1}{T_1} = \dfrac{E_2}{T_2}$:
$\dfrac{E_0}{300} = \dfrac{E_2}{450}$
$E_2 = E_0 \times \dfrac{450}{300} = \dfrac{3E_0}{2}$
Answer: $3E_0 / 2$
$E \propto T$
Initial temperature $T_1 = 27^\circ\text{C} = 27 + 273 = 300\text{ K}$
Final temperature $T_2 = 177^\circ\text{C} = 177 + 273 = 450\text{ K}$
Using the relation $\dfrac{E_1}{T_1} = \dfrac{E_2}{T_2}$:
$\dfrac{E_0}{300} = \dfrac{E_2}{450}$
$E_2 = E_0 \times \dfrac{450}{300} = \dfrac{3E_0}{2}$
Answer: $3E_0 / 2$
Hence, the correct answer is option: 'C'.