Thermal Properties of Matter

JEE Main level, single correct, medium

62 of 126

Question 62 of 126

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^{\circ}C will be (Given gas constant R=8.3JK−1mol−1R=8.3J{K}^{-1}{mol}^{-1})

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Answer: (C) 498J498J

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=nCvΔT=(2)(32R)(20)=3×8.3×20=498JQ=nCv\Delta T=\left(2\right)\left(\dfrac{3}{2}R\right)\left(20\right)=3\times 8.3\times 20=498J

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