91影视

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 ${\mathrm{P}}_{\mathrm{a}}=0.50鈥�\mathrm{atm}$and volume $V_a=0.20鈥�L$

At point b: pressure $P_b=2.0鈥�atm$ and volume $V_b=0.20鈥�L$

At point c: pressure $P_b=2.0鈥�atm$and volume $V_c=?$

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

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,

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

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 $0.80\mathrm{L}$.

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,

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

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.

The segment wise heat energies are calculated below,

For the segment ab:

$\mathrm{Q}={\mathrm{nC}}_{\mathrm{v}}\mathrm{螖罢}$

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:

$Q=n{C}_p螖T$

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.

The segment wise heat energies are calculated below,

For the segment ab:

$螖U=n{C}_v螖T$

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.

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

The internal energy is increased.

For the segment bc:

The temperature remains constant, $螖T=0$

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

For the segment ca:

$\mathrm{螖鲍}={\mathrm{nC}}_{\mathrm{v}}\mathrm{螖罢}$

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.

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

The internal energy is decreased.