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.

Answer 1

The change in the internal energy of the gas 36.12 J.

Explanation:

Step 1:

Given data in the question

gas Mass, m = 5 g

Change in system temperature, ∆T = 25 − 15°C = 10°C

constant volume  Specific heat $\mathrm{C}_{v}=0.172 \mathrm{cal} / \mathrm{g}-^{\circ} \mathrm{C}$

Mechanical Heat Equivalent, J = 4.2 J/cal

Step 2:

From the thermodynamics first law,

dQ =  dU + dW

Now,

ΔV = 0 (Rigid container wall keeping volume constant)

So,  

dW = PΔV = 0

Step 3:

Therefore,  

dQ = dU  (From the first law)

$Q=m c_{v} d T$

= 5 × 0.172 × 10

= 8.6 cal = 8.6 × 4.2 J

= 36.12 J

Therefore, adjustment of the system's internal energy is 36.12 J.

Answer 2

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