Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 3

Question

Consider the expansion of 60.0 moles of a monatomic ideal gas along processes 3 and 4 in the figure. On process 3 the gas expands at constant pressure from an initial volume of 1.00 m^3 to a final volume of 3.00 m^3. On process 4 the gas is heated at constant volume from an initial pressure of 106 kPa to a final pressure of 212 kPa.

Part A

How much heat is added to the gas during these two processes (process 3 and 4)?

Part B

How much work does the gas do during this expansion?

Part C

What is the change in the internal energy of the gas?

Explanation / Answer

(a) Process 3 is a constant pressure process. For such a process the heat added to a system equals its change in enthalpy.
Q = H
The enthalpy of an ideal gas is given by:
H = nCpT
So the change in enthalpy for a constant amount is:
H = nCpT
For a monatomic ideal gas, which has molar heat capacity at constant volume of
Cp = (5/2)R
it is
Q = H = (5/2)nRT

Using ideal gas you can express Q in terms of pressure and volume
nRT = pV
=>
nRT = (pV) = pV (because p is constant)

Hence,
Q = H = (5/2)pV
= (5/2) p_i(V_f - V_i)
= (5/2) 106 kPa (3.0 m³ - 1.0 m³)
= 530 kJ


Process 4 is a constant volume process. For such a process the heat added to a system equals its change in in internal energy.
Q = U
With
U = nCvT
and
Cv = (3/2)R
for a monatomic ideal gas follows
Q = U = (3/2)nRT
From ideal gas law follows that
nRT = (pV) = Vp
because V is constant in this process
Hence,
Q = U = (3/2)Vp
= (3/2) V_f(p_f - p_i)
= (3/2) 3.0 m³ (212 kPa - 106 kPa)
= 477 kJ

(b) Work done to surrounding is given by:
W = p dV form initial to final volume

In process 4 the volume is constant so no work is done

For constant pressure process 3 the work integral simplifies to:
W = p dV = pV
= p_i (V_f - V_i)
= 106 kPa (3.0 m³ - 1.0 m³)
= 212 kJ

(c) Total change in internal energy equals total heat added to the gas minus the work it does to the surrounding:
U = Q - W
= 530 kJ + 477 kJ - 212 kJ
= 795 kJ

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