Question

When 0.1 mole of a gas absorbs 41.75j of heat at constant volume the rise in temperature occurs equal to 20 degrees.The gas must be?

Answer 1

Explanation:

n = 0.1 mole, dt = 20℃, q = 41.75 J

R = 8.314 J/mol°k

we know that heat supplied at constant volume is

= n*Cv*dt

41.75 = 0.1*Cv*20

Cv = 20.875

but Cv = R/(gamma - 1)

therefore

R/(gamma - 1) = 20.875

gamma - 1 = 8.314/20.875

gamma = 1 + 0.4 = 1.4

we know that specific heat ratio gamma is having value of 1.4 for diatomic gas.

hence the gas must be diatomic.