KVPY 2019 SB SX Exam Previous Papers

Question : 32 of 120

Marks: +1, -0

0.02 moles of an ideal diatomic gas with initial temperature 20°C is compressed from 1500 cm

^{3}

to 500 cm

^{3}

. The thermodynamics process is such that

PV^2 = β

where β is a constant. Then the value of β is close to (The gas constant, R = 8.31 J/K/mol)

Solution:

PV^2 = B

= constant
& PV = nRT
⇒ nRTV = β = costant
from initial condition of T & V
β = 0.073 pa – m

^{6}

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