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JEE Main 202229 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
29 June
Shift
Morning Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2022 · 29 June · Morning Shift · Physics · Thermal Properties of Matter

A cylinder of fixed capacity of 44 . 8 litres contains helium gas...

A cylinder of fixed capacity of 44.844 . 8 litres contains helium gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0°C20 . 0 ° C will be (Given gas constant R=8.3JK−1mol−1R = 8 . 3J{K}^{- 1}{mol}^{- 1})

Options

  1. A249J249J
  2. B415J415J
  3. C498J498J
  4. D830J830J

Correct answer

C. 498J498J

Step-by-step solution

At STP volume occupied by one mole of a gas is 22.4L22 . 4L. Therefore, number of moles are, n=44.822.4=2moln = \dfrac{44 . 8}{22 . 4} = 2mol.

The process is isochoric process as the volume of the cylinder remains fixed and helium is a monatomic gas. Therefore, amount of heat needed will be,

Q=nCv\DeltaT=(2)(32R)(20)=3×8.3×20=498JQ = n Cv \DeltaT = \left(2\right)\left(\dfrac{3}{2}R\right)\left(20\right) = 3 \times 8 . 3 \times 20 = 498J

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