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=CV=The relationship for heat transfer in terms of
Cp is:
Q=nCp∆TSubstituting in known values:
40=n⋅⋅∆TFrom 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⋅R∆T∆U=⋅nR∆TUsing Equation (i), we substitute for
nR∆T :
∆U=×16This simplifies to:
∆U=24JThus, the increase in the internal energy of the gas is 24 J .
