Kinds of Thermodynamic Processes, Internal Energy of an Ideal Gas, and Heat Capacities of an Ideal Gas Problems and Solutions 2

Q#19.21

In an experiment to simulate conditions inside an automobile engine, 0.185 mol of air at a temperature of 780 K and a pressure of 3.00 $\times 10^6$ Pa is contained in a cylinder of volume 40.0 cm$^3$.Then 645 J of heat is transferred to the cylinder. 

(a) If the volume of the cylinder is constant while the heat is added, what is the final temperature of the air? Assume that the air is essentially nitrogen gas, and use the data in Table 19.1 even though the pressure is not low. Draw a pV-diagram for this process. 

(b) If instead the volume of the cylinder is allowed to increase while the pressure remains constant, find the final temperature of the air. Draw a pV-diagram for this process. 

Answer:

For constant volume  Q = n$C_v\Delta T$

For constant pressure Q = n$C_p\Delta T$

Given: For helium, $C_v$ = 20.78 J/mol.K and $C_p$ = 29.07 J/mol.K

(a) Using Eq.  Q = n$C_v\Delta T$

$\Delta T = \frac{Q}{nC_v}=\frac{648 \ J}{(0.185 \ mol)(20.76 \ J/mol.K)} = 167.9 \ K$

$\Delta T = T_2 - T_1$

167.9 K = $T_2$ - 780 K

$T_2$ = 948 K

The pV-diagram is sketched in Figure 19.21a.

(b) Using Eq.  Q = n$C_v\Delta T$

$\Delta T = \frac{Q}{nC_p}=\frac{648 \ J}{(0.185 \ mol)(29.07 \ J/mol.K)} = 119.9 \ K$

$\Delta T = T_2 - T_1$

119.9 K = $T_2$ - 780 K

$T_2$ = 900 K

The pV-diagram is sketched in Figure 19.21b.

At constant pressure some of the heat energy added to the gas leaves the gas as expansion work and the internal energy change is less than if the same amount of heat energy is added at constant volume. ΔT is proportional to ΔU.

Fig.21

Q#19.22

When a quantity of monatomic ideal gas expands at a constant pressure of 4.00 $\times 10^4$ Pa, the volume of the gas increases from 2.00 $\times 10^{-3} \ m^3$ to 8.00 $\times 10^{-3} \ m^3$. What is the change in the internal energy of the gas? 

Answer:

For an ideal gas, , ΔU = $C_v\Delta T$ and at constant pressure, pΔV = nRΔT.

$C_v=\frac{3}{2}R$,  for a monatomic gas.

ΔU = n$\left(\frac{3}{2}R\right)\Delta T=\frac{3}{2}p\Delta V$

ΔU =\frac{3}{2}(4.00 \times 10^4 \ Pa)$ (8.00 $\times 10^{-3} \ m^3-$ 2.00 $\times 10^{-3} \ m^3$)

ΔU = 360 J

W = nR$\Delta T$ = $\frac{2}{3}\Delta U=\frac{2}{3} \times 360 \ J$

W = 240 J

Q = n$C_p\Delta T$ = $n\left(\frac{5}{2}R\right)\Delta T=\frac{5}{2}\Delta U$

Q = 600 J.

600 J of heat energy flows into the gas. 240 J leaves as expansion work and 360 J remains in the gas as an increase in internal energy. 

Q#19.23

Heat Q flows into a monatomic ideal gas, and the volume increases while the pressure is kept constant. What fraction of the heat energy is used to do the expansion work of the gas?

Answer:

ΔU = Q - W. For an ideal gas, , ΔU = $C_v\Delta T$ and at constant pressure, pΔV = nRΔT.

$C_v=\frac{3}{2}R$,  for a monatomic gas.

ΔU = n$\left(\frac{3}{2}R\right)\Delta T=\frac{3}{2}p\Delta V=\frac{3}{2}W$

Then ΔU = Q - W, So

Q = ΔU + W = $\frac{3}{2}W$ + W 

Q =  $\frac{5}{2}W$

So, $\frac{W}{Q}=\frac{2}{5}$

For diatomic or polyatomic gases, $C_v$ is a different multiple of R and the fraction of Q that is used for expansion work is different. 

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