Chapter 15, Problem 11P

(a)

To determine

The temperature and pressure at point C.

Answer to Problem 11P

The temperature and pressure at point C are $\text{1180}\text{K}$ and $98.0\text{kPa}$ respectively.

Explanation of Solution

Write the expression for temperature using Ideal gas equation.

$T=\frac{PV}{nR}$

Here, T is the temperature, P is the pressure, V is the volume, n is the number of moles and R is the gas constant.

From the graph given in the question, the pressure at point C is 98.0 kPa.

Conclusion:

Substitute 98.0 kPa for P, 2.00 L for V, 0.0200 mol for n and $8.314{\text{Jmol}}^{-1}{\text{K}}^{-1}$ for R to get T.

$\begin{array}{c}T=\frac{\left(98.0\text{kPa}\right)\left(2.00\text{L}\right)}{\left(0.0200\text{mol}\right)\left(8.314{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)}\\ =\frac{\left(98.0\text{kPa}\right)\left(\frac{{10}^3\text{Pa}}{1\text{kPa}}\right)\left(2.00\text{L}\right)\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}}{1\text{L}}\right)}{\left(0.0200\text{mol}\right)\left(8.314{\text{Jmol}}^{-1}{\text{K}}^{-1}\right)}\\ \approx \text{1180}\text{K}\end{array}$

Therefore, temperature and pressure at point C are $\text{1180}\text{K}$ and $98.0\text{kPa}$ respectively.

(b)

To determine

The change in internal energy of the gas between points A and B.

Answer to Problem 11P

The change in internal energy of the gas between points A and B is $-\text{200}\text{J}$.

Explanation of Solution

Write the expression for change in internal energy.

$\Delta U=\frac{3}{2}nR\Delta T$

Here, $\Delta U$ is the change in internal energy and $\Delta T$ is the change intemperature.

As the volume is constant, the ideal gas equation can be written as,

$V\Delta P=nR\Delta T$

Replace $nR\Delta T$ by $V\Delta P$ in the expression for $\Delta U$.

$\begin{array}{c}\Delta U=\frac{3}{2}V\Delta P\\ =\frac{3}{2}V\left(P_B-P_A\right)\end{array}$

Here, $P_A$ is the pressure at point A and $P_B$ is the pressure at point B.

Conclusion:

Substitute 98.0 kPa for $P_A$, 1.00 L for V and 230 kPa for $P_B$ to get $\Delta U$.

$\begin{array}{c}\Delta U=\frac{3}{2}\left(1.00\text{L}\right)\left(98.0\text{kPa}-230\text{kPa}\right)\\ =\frac{3}{2}\left(1.00\text{L}\right)\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}}{1\text{L}}\right)\left(98.0\text{Pa}-230\text{Pa}\right)\left(\frac{{10}^3\text{Pa}}{1\text{kPa}}\right)\\ \approx -\text{200}\text{J}\end{array}$

Therefore, change in internal energy of the gas between points A and B is $-\text{200}\text{J}$.

(c)

To determine

The work done by the gas.

Answer to Problem 11P

The work done by the gas is $\text{66}\text{J}$.

Explanation of Solution

Work done by the gas is equal to the area under the curve. Therefore, work done is equal to the area of triangle ABC. Write the expression for area of triangle ABC.

$W=\frac{1}{2}\left(AB\right)\left(BC\right)$

The length AB is,

$\begin{array}{c}AB=230\text{kPa}-98.0\text{kPa}\\ =132\text{kPa}\end{array}$

The length BC is,

$\begin{array}{c}BC=2.00\text{L}-\text{1}\text{.00}\text{L}\\ =\text{1}\text{.00}\text{L}\end{array}$

Conclusion:

Substitute 132 kPa for AB and 1.00 L for BC to get W.

$\begin{array}{c}W=\frac{1}{2}\left(132\text{kPa}\right)\left(1.00\text{L}\right)\\ =\frac{1}{2}\left(132\text{kPa}\right)\left(\frac{{10}^3\text{Pa}}{1\text{kPa}}\right)\left(1.00\text{L}\right)\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}}{1\text{L}}\right)\\ =\text{66}\text{J}\end{array}$

Therefore, work done by the gas is $\text{66}\text{J}$.

(d)

To determine

The total change in internal energy of the gas for one cycle.

Answer to Problem 11P

The total change in internal energy of the gas for one cycle is $\text{0}\text{J}$.

Explanation of Solution

Write the expression for change in internal energy.

$\Delta U=\frac{3}{2}nR\Delta T$

Here, $\Delta U$ is the change in internal energy and $\Delta T$ is the change in temperature.

For one complete cycle, the intial and final temperatures are the same. Therefore, change in temperature is zero. As a result, total change in internal energy of the gas is zero.

Conclusion:

Therefore, total change in internal energy of the gas for one cycle is $\text{0}\text{J}$.