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Question
5 g of a gas is contained in a rigid container and is heated from 15°C to 25°C. Specific heat capacity of the gas at constant volume is 0.172 cal g−1 °C−1 and the mechanical equivalent of heat is 4.2 J cal−1. Calculate the change in the internal energy of the gas
Solution
Given:
Mass of the gas, m = 5 g
Change in temperature of the system, ∆T = 25 − 15°C = 10°C
Specific heat at constant volume, Cv = 0.172 cal/g -°C
Mechanical equivalent, J = 4.2 J/cal
From the first law of thermodynamics,
dQ = dU +dW
Now ,
ΔV = 0 (Rigid wall of the container keeps the volume constant)
So, dW =PΔV =0
Therefore,
dQ =dU (From the first law)
Q = mcvdT = 5×0.172×10
= 8.6 cal = 8.6 × 4.2 J
= 36.12 J
So, change in internal energy of the system is 36.12 J.
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