Question

With the pressure held constant at 210 kPa , 49 mol of a monatomic ideal gas expands from an initial volume of 0.60 m3 to a final volume of 1.7 m3 .

A) How much work was done by the gas during the expansion?

B) What were the initial temperature of the gas?

C) What were the final temperature of the gas?

D) What was the change in the internal energy of the gas?

E) How much heat was added to the gas?

Answer 1

B) What were the initial temperature of the gas?

The process is an isobaric heating of the gas.

Initial temp of gas can be found with ideal gas law:
P1*V1 = n*R*T1

We know n, V1, and P1. R is a constant independent of flavor.
Solve for T1:
T1 = P1*V1/(n*R)

Plug in data:
P1 := 210,000 Pa; V1:= 0.60m^3; n:=49 moles; R := 8.314 J/mole-K

Therefore,
T1 = 309 Kelvin

C) What were the final temperature of the gas?

As for the final temperature, we know the final volume and final pressure, therefore, the same equation form applies. Amount of gas is unchanged, since it is a closed system.
T2 = P2*V2/(n*R)

Data:
P2 := 210,000 Pa; V2:= 1.7 m^3; n:=49 moles; R := 8.314 J/mole-K;

Result:
T2 = 876 Kelvin

By definition of adiabatic index,
k = cp/cv

For an ideal gas, the two molar specific heats have a common difference
cp = cv + R

This is because enthalpy and internal energy are related by
h = u + R*T

Solving for the molar isochoric specific heat:
cv = R/(k - 1)

Now, to find the change in internal energy,
deltaU = n*cv*(T2 - T2)

Substitute:
deltaU = n*R/(k - 1)*(P2*V2/(n*R) - P1*V1/(n*R))

Simplify:
deltaU = (P2*V2 - P1*V1)/(k - 1)

Since isobaric, P1 = P2:
deltaU = P1*(V2 - V1)/(k - 1)

Plug in data:
P1:=210000 Pa; V2:=1.7 m^3; V1:=0.60 m^3; k:=5/3; (Because it is a monatomic gas, this means that the adiabatic index is k = 5/3)

Result:
deltaU = 346,500 Joules

Heat added to the gas:
Use the 1st law equation relating heat added, increase in U, and work done

Q = deltaU + W

We know deltaU. Work is done, because the gas expands.

Find work using the isobaric version of ideal gas work:
W = P1*(V2 - V1)

Substitute with deltaU to find expression for Q:
Q = P1*(V2 - V1)/(k - 1) + P1*(V2 - V1)

Simplfiy:
Q = P1*(V2 - V1)*k/(k - 1)

Note, since isobaric molar specific heat is the following, the enthalpy shortcut can also tell us the heat added to the gas without having to think about work.
cp = k*R/(k - 1)

Plug in same set of data:
Q =581000 Joules