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Kinetic Theory · Question 18

JEE Main 2025 (Online) 29th January Morning ShiftNumericalEasy📈 Topic in 23% papers+4
1

A container of fixed volume contains a gas at 27 . To double the pressure of the gas, the temperature of gas should be raised to .

Answer

327

Solution

At constant volume: P1T1 = P2T2

T1 = 27 + 273 = 300 K, P2 = 2P1

T2 = 2T1 = 600 K

T2 = 600 - 273 = 327

Answer: 327

JEE Main 2025 (Online) 29th January Morning Shift

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2

For a particular ideal gas which of the following graphs represents the variation of mean squared velocity of the gas molecules with temperature?

  1. [figure]
  2. [figure]
  3. [figure]
  4. [figure]
JEE Main 2025 (Online) 28th January Morning Shift
3

The ratio of vapour densities of two gases at the same temperature is 425, then the ratio of r.m.s. velocities will be :

  1. 52
  2. 254
  3. 425
  4. 25
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4

The kinetic energy of translation of the molecules in 50 g of CO2 gas at 17 is :

  1. 4205.5 J
  2. 3582.7 J
  3. 3986.3 J
  4. 4102.8 J
JEE Main 2025 (Online) 28th January Evening Shift

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