A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature 50°C. The gas is now heated to double its temperature. The modified pressure is ____ Pa.

Numerical Answer Type:
Enter a numerical value Answer: 3730 to 3730 +4 marks

Solution & Explanation

Related Formula
P ∝ T (Gay-Lussac's Law for constant V)
Core Logic

Since the gas is filled in a closed cylinder, the volume V remains constant. Therefore, P ∝ T (where T must be in Kelvin). Initial temperature Tᵢ = 50°C = 273 + 50 = 323 K. Final temperature Tf = 100°C = 273 + 100 = 373 K.

Note: The phrasing "double its temperature" is slightly ambiguous (Celsius vs Kelvin). As per standard interpretations in such exams, "double its temperature" given in Celsius means 100°C (which is 373 K). If it meant doubling absolute temperature, final would be 646 K. The official interpretation used here doubles the Celsius scale value.

Step 1: Calculate Final Pressure

Using the relation:

(Pf)/(Pᵢ) = (Tf)/(Tᵢ) (Pf)/(3.23 × 10³) = (373)/(323) Pf = 3.23 × 10³ × (373)/(323) = 3.73 × 10³ Pa Pf = 3730 Pa
Pattern Recognition

Be highly alert to 'double temperature' traps when given in Celsius. The question specifically intends 50^° arrow 100^°, not 323K arrow 646K. Calculating with 373K nicely cancels with the 3.23kPa.

Chapter Mix

Class 11 Physics: Kinetic Theory Class 11 Physics: Thermodynamics

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Kinetic Theory

More Kinetic Theory Previous-Year Questions — Page 6

Q31 jee_main_2024_31_jan_morning Kinetic Energy Of Gas Molecules
The parameter that remains the same for molecules of all gases at a given temperature is :
  • A. kinetic energy
  • B. momentum
  • C. mass
  • D. speed

Solution

Related Formula
KE = (f)/(2)kT
Core Logic

The average translational kinetic energy of any gas molecule depends only on the absolute temperature of the gas and is independent of the nature or mass of the gas.

For 1 mole of any ideal gas, the average translational kinetic energy is (3)/(2)RT. Therefore, at a given temperature, the kinetic energy parameter is uniform across all ideal gases.

Pattern Recognition

Temperature is directly proportional to average translational kinetic energy. If T is constant, KE is constant for all gases regardless of mass.

Chapter Mix

Class 11 Physics: Kinetic Theory

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