Heat and Thermodynamics

To analyze the increase in internal energy when heat is absorbed by a monoatomic gas:
First, we know that the heat absorbed, Q, is given as 40 J .
For a monoatomic ideal gas, the specific heat capacities are:
‌Cp=‌

5R

2

‌CV=‌

3R

2

The relationship for heat transfer in terms of Cp is:
Q=nCp∆T
Substituting in known values:
40=n⋅‌

5R

2

⋅∆T
From this, we obtain:
nR∆T=16‌‌(i)
Next, we determine the increase in internal energy, ∆U, which is given by:
‌∆U=nCV∆T
‌∆U=n⋅‌

3

2

R∆T
‌∆U=‌

3

2

⋅nR∆T
Using Equation (i), we substitute for nR∆T :
∆U=‌

3

2

×16
This simplifies to:
∆U=24J
Thus, the increase in the internal energy of the gas is 24 J .