Physics Q100 – AP-EAMCET 2025

Home Physics Kinetic TheoryQuestion #204

A closed vessel contains a gas at a pressure P. If 50% of the mass of the gas is removed and rms speed of the gas molecules is increased by 20%, then the pressure of the remaining gas is

Clue 1 of 3: Ideal Gas Law and Kinetic Theory

Relate pressure to the mass of the gas and the root mean square (rms) speed of its molecules.

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Step 1: Relate Pressure to Mass and RMS Speed✦ Active

Start with the ideal gas equation $P V = n R T$ . Recall the relation between temperature and rms speed: $v_{r m s} = \sqrt{\frac{3 R T}{M}}$ , which implies $T = \frac{M v_{r m s}^{2}}{3 R}$ . Substitute $n = \frac{m}{M}$ (where $m$ is the total mass of the gas and $M$ is its molar mass) into the ideal gas equation and then substitute the expression for $T$ .

P V = \frac{m}{M} R T

⟹ P = \frac{m}{M V} R T

v_{r m s} = \sqrt{\frac{3 R T}{M}} ⟹ T = \frac{M v_{r m s}^{2}}{3 R}

P = \frac{m}{M V} R (\frac{M v_{r m s}^{2}}{3 R}) = \frac{m v_{r m s}^{2}}{3 V}

💡 Teacher's Secret Hint

This derived formula $P = \frac{m v_{r m s}^{2}}{3 V}$ is crucial for problems involving changes in mass and rms speed while the volume is constant.

Step 2: Define Initial Conditions○ Expand

Let the initial mass of the gas be $m_{1}$ , initial rms speed be $v_{r m s, 1}$ , and initial pressure be $P_{1} = P$ . The volume of the vessel is $V$ and remains constant.

P_{1} = \frac{m_{1} v_{r m s, 1}^{2}}{3 V}

Step 3: Define Final Conditions○ Expand

50\% of the mass is removed, so the new mass $m_{2}$ is $m_{1} - 0.5 m_{1} = 0.5 m_{1} = \frac{1}{2} m_{1}$ . The rms speed is increased by 20\%, so the new rms speed $v_{r m s, 2}$ is $v_{r m s, 1} + 0.20 v_{r m s, 1} = 1.2 v_{r m s, 1}$ . The volume $V$ remains constant, so $V_{2} = V$ .

m_{2} = \frac{1}{2} m_{1}

v_{r m s, 2} = 1.2 v_{r m s, 1}

V_{2} = V

Step 4: Calculate the Final Pressure○ Expand

Substitute the new mass and rms speed into the pressure formula derived in Step 1 to find the final pressure $P_{2}$ .

P_{2} = \frac{m_{2} v_{r m s, 2}^{2}}{3 V_{2}} = \frac{(\frac{1}{2} m_{1}) (1.2 v_{r m s, 1})^{2}}{3 V}

P_{2} = \frac{\frac{1}{2} m_{1} (1.44 v_{r m s, 1}^{2})}{3 V} = \frac{1.44}{2} (\frac{m_{1} v_{r m s, 1}^{2}}{3 V})

P_{2} = 0.72 P_{1}

Step 5: Express Final Pressure as a Fraction of Initial Pressure○ Expand

Convert the decimal value of $0.72$ into a fraction.

P_{2} = 0.72 P = \frac{72}{100} P = \frac{18}{25} P

💡 Teacher's Secret Hint

Always simplify fractions to their lowest terms to match the options provided.

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Physics Question 100 – AP-EAMCET 2025

Clue 1 of 3: Ideal Gas Law and Kinetic Theory

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