Question

Figure P19.43 shows a pV-diagram for 0.0040 mol of ideal ${\mathbf{H}}_{\mathbf{2}}$ gas. The temperature of the gas does not change during segment bc. (a) What volume does this gas occupy at point c? (b) Find the temperature of the gas at points a, b, and c. (c) How much heat went into or out of the gas during segments ab, ca, and bc? Indicate whether the heat has gone into or out of the gas. (d) Find the change in the internal energy of this hydrogen during segments ab, bc, and ca. Indicate whether the internal energy increased or decreased during each segment.

Short answer

(a) The volume of the gas at point c is .

(b) The temperature of the gas at points a, b, c are respectively.

(c) The amount of heat in segment:

ab: +76 J

bc: + 56 J

ca: -107. J

(d) The change in internal energy of the hydrogen in segment:

ab: +7.6 J

bc: 0 J

bc: -76 J

Step-by-step solution

Identification of the given data.

A gas that perfectly complies with the ideal-gas law is one in which there is no molecular attraction.

Consider the given data as below.

Number of moles of the gas is n =0.0040 mol.

Universal gas constant $\mathrm{R}=8.315\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}$.

From the P-V diagram,

At point a: pressure ${\mathrm{P}}_{\mathrm{a}}=0.50\hspace{0.33em}\mathrm{atm}$and volume $V_a=0.20\hspace{0.33em}L$

At point b: pressure $P_b=2.0\hspace{0.33em}atm$ and volume $V_b=0.20\hspace{0.33em}L$

At point c: pressure $P_b=2.0\hspace{0.33em}atm$and volume $V_c=?$

(a) Determine the volume at point c.

For segment bc, the process is isothermal, i.e. $\Delta T=0$.

This implies, $T_b=T_c$. So,

According to the ideal gas equation,

$PV=RT$

Here,

At constant temperature, the product of pressure temperature is constant as R is the universal gas constant. Thus,

$$\begin{gathered} P=\frac{RT}{V} \\ V\propto \frac{1}{P} \end{gathered}$$

Solve for volume at point c,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{c}}=\left({\mathrm{V}}_{\mathrm{b}}\right)\left(\frac{{\mathrm{p}}_{\mathrm{b}}}{{\mathrm{P}}_{\mathrm{c}}}\right) \\ =\left(0.2\mathrm{L}\right)\left(\frac{2.0\mathrm{atm}}{0.50\mathrm{atm}}\right) \\ =0.80\mathrm{L} \end{gathered}$$

Thus, the volume at point c is $0.80\mathrm{L}$.

(b) Determine temperature of the gas at points a, b, c.

The temperature at point a from the ideal gas equation is,

$T_a=\frac{P_a{V}_a}{nR}$

Substitute all the values in the above equation,

$$\begin{gathered} {\mathrm{T}}_{\mathrm{a}}=\frac{\left(0.50\mathrm{atm}\right)\left(1.013\times {10}^5{\displaystyle \frac{\mathrm{Pa}}{\mathrm{atm}}}\right)\left(0.20\times {10}^{-3}{\mathrm{m}}^3\right)}{\left(0.00040\mathrm{mol}\right)\left(8.315{\displaystyle \frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}}\right)} \\ =305\mathrm{K} \end{gathered}$$

Since ab is isochoric process so,

$T_b=T_a\frac{P_b}{P_a}$

Also from part (a), it has been established that $T_b=T_c$.

So,

$T_b=T_c$

$$\begin{gathered} =\left(305\mathrm{K}\right)\left(\frac{2.0\mathrm{atm}}{0.50\mathrm{atm}}\right) \\ =1220\mathrm{K} \end{gathered}$$

Thus, the temperatures at point a, b, c are 305K, 1220 K, 1220 K respectively.

(c) Determine the amount of heat in the segments ab, bc, ca.

The segment wise heat energies are calculated below,

For the segment ab:

$\mathrm{Q}={\mathrm{nC}}_{\mathrm{v}}\mathrm{\Delta T}$

Substitute $0.0040\mathrm{mol}$ for n, $\frac{5}{2}$for $C_v$, and 8.315$\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}$ for in the above equation.

$$\begin{gathered} \mathrm{Q}=\left(0.0040\mathrm{mol}\right)\left(\frac{5}{2}\right)\left(8.315\frac{\mathrm{j}}{\mathrm{mol}.\mathrm{K}}\right)\left(1220\mathrm{K}-305\mathrm{K}\right) \\ =+76\mathrm{J} \end{gathered}$$

The heat in this segment is positive and is added to the gas.

For the segment bc:

$\begin{array}{l}\mathrm{Q}=\mathrm{W}\\ =\mathrm{nR}\mathrm{T}\ln \left(\frac{{\mathrm{v}}_{\mathrm{e}}}{{\mathrm{V}}_{\mathrm{b}}}\right)\end{array}$

Substitute 0.0040 mol for n and role="math" localid="1668086834025" $8.315\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}$ for in the above equation.

$$\begin{gathered} Q=\left(0.0040\mathrm{mol}\right)\left(8.315\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}\right)\left(1220\mathrm{K}\right)\mathrm{In}\left(\frac{0.80\mathrm{L}}{0.20\mathrm{L}}\right) \\ =+56\mathrm{J} \end{gathered}$$

The heat in this segment is positive and is added to the gas.

For the segment ca:

$Q=n{C}_p\Delta T$

Substitute 0.0040 mol for n , $\frac{7}{2}$ for $C_p$, and $\left(8.315\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}\right)$for in the above equation.

$$\begin{gathered} Q=\left(0.0040\mathrm{mol}\right)\left(\frac{7}{2}\right)\left(8.315\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}\right)\left(305\mathrm{K}-1220\mathrm{K}\right) \\ =-107\mathrm{J} \end{gathered}$$

The heat in this segment is negative and is released by the gas.

(d) Determine the change in internal energy in segments ab, bc, ca.

The segment wise heat energies are calculated below,

For the segment ab:

$\Delta U=n{C}_v\Delta T$

Substitute $0.0040mol$for n , $\frac{5}{2}$for $C_v$, and $8.315\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}$ for R in the above equation.

$$\begin{gathered} ∆U=\left(0.0040\mathrm{mol}\right)\left(\frac{5}{2}\right)\left(8.315\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}\right)\left(1220\mathrm{K}-305\mathrm{K}\right) \\ =+76\mathrm{J} \end{gathered}$$

The internal energy is increased.

For the segment bc:

The temperature remains constant, $\Delta T=0$

So, $\mathrm{\Delta U}=0$, i.e. the internal energy remains constant.

For the segment ca:

$\mathrm{\Delta U}={\mathrm{nC}}_{\mathrm{v}}\mathrm{\Delta T}$

Substitute 0.0040 mol for n , $\frac{5}{2}$ for $C_v$, and 8.315 $\frac{J}{mol.k}$for R in the above equation.

$$\begin{gathered} ∆\mathrm{U}=\left(0.0040\mathrm{mol}\right)\left(\frac{5}{2}\right)\left(8.315\frac{\mathrm{J}}{\mathrm{mol}.\mathrm{K}}\right)\left(305\mathrm{K}-1220\mathrm{K}\right) \\ =-76\mathrm{J} \end{gathered}$$

The internal energy is decreased.