Test Index

MHTCET 23 Apr 2025 Shift 2 Paper

Question : 5 of 150

Marks: +1, -0

Select the correct statement.

The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules.
The temperature of gas is $-73^{\circ} C$ . When the gas is heated to $527^{\circ} C$ , the r.m.s. speed of the molecules is doubled.
The temperature of gas is $-100^{\circ} C$ . When the gas is heated to $+627^{\circ} C$ , the r.m.s. speed of the molecules is four times.
The product of pressure and volume of an ideal gas will be equal to half the translational kinetic energy.

Solution:

Concept:

For an ideal gas, translational kinetic energy is

\frac{3}{2}PV

, and rms speed depends on the square root of absolute temperature.

Explanation:

The translational kinetic energy of an ideal gas is given by:

\text{K.E.} = \frac{3}{2}PV

Therefore, option (A) saying

PV

equals translational kinetic energy is incorrect.

Option (D) saying

PV

equals half the translational kinetic energy is also incorrect.

Now check the rms speed relation:

v_{\text{rms}} = \sqrt{\frac{3RT}{M}}

v_{\text{rms}} \propto \sqrt{T}

where

T

is absolute temperature in kelvin.

For option (B):

T_1 = -73^\circ C = 200\,K

and

T_2 = 527^\circ C = 800\,K

\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{800}{200}} = \sqrt{4} = 2

Thus

v_2 = 2v_1

, so the rms speed is doubled. Option (B) is correct.

For option (C):

T_1 = -100^\circ C = 173\,K

and

T_2 = 627^\circ C = 900\,K

\frac{v_2}{v_1} = \sqrt{\frac{900}{173}} \approx 2.28

The speed is not four times, so option (C) is incorrect.

Answer:

The correct statement is option (B).

Go to Question:

Prev Question

Next Question