Chapter 20, Problem 52SP

Figure 20-6 is the $P–V$ diagram for 25.0 g of an enclosed ideal gas. At A the gas temperature is 200 °C. The value of $c_{\upsilon }$ for the gas is $0.150\text{ cal/g}\cdot \text{°C}$. (a) What is the temperature of the gas at B? (b) Find $\Delta U$ for the portion of the cycle from A to B. (c) Find $\Delta W$ for this same portion. (d) Find $\Delta Q$ for this same portion.

To determine

The temperature of the gas at B, if the temperature at A is $200°\text{C}$, as shown in Fig. 20-6.

Answer to Problem 52SP

Solution:

$1.42\times {10}^3\text{ K}$

Explanation of Solution

Given data:

Refer to Fig. 20-6.

The temperature at A is $400°\text{C}$.

The mass of the ideal gas enclosed is $25.0\text{ g}$.

The specific heat of the gas at constant volume is $0.150\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$

The gas follows the process A to B in the thermodynamic cycle shown in Fig. 20-6.

Formula used:

The gas equation between initial and final condition of a gas can be written as,

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

Here, $P_1$ represent the initial pressure, $V_1$ represents the initial volume, $T_1$ represents the initial temperature, $P_2$ represent the final pressure, $V_2$ represents the final volume, and $T_2$ represents the final temperature.

The formula for the conversion of the initial temperature of a gas from the Celsius scale to the Kelvin scale is,

$T_1=\left[t_1+273\right]\text{ K}$

Here, $t_1$ is the initial temperature of the gas in Celsius.

The area of a trapezium is calculated by the formula,

$A=\frac{1}{2}\left(a+b\right)\left(h\right)$

Here, $A$ is the area of the trapezium, $a$ and $b$ are the lengths of the parallel sides of the trapezium, $a+b$ represents the sum of the parallel sides, and $h$ is the perpendicular distance between the parallel sides of the trapezium.

The work done in a thermodynamic process is given by the area under the line representing the process in the pressure-volume diagram.

$W=A_{PVD}$

Here, $A_{PVD}$ is the area under the line representing the process in the pressure-volume diagram and $W$ is the work done in the process.

Explanation:

Draw the thermodynamic cycle diagram given in Fig- 20.6.

Recall the expression for the conversion of temperature at A from Celsius to Kelvin.

$T_A=t_A+273$

Here, $T_A$ is the temperature at point A in Kelvin and $t_A$ is the temperature at point A in Celsius.

Substitute $200°\text{C}$ for $t_A$

$\begin{array}{c}{T}_A=200°\text{C}+273\\ =473\text{ K}\end{array}$

Refer to the diagram and write the values of pressure and volume at points A and B, respectively.

$\begin{array}{l}{P}_A=6\text{ atm}\\ P_B=4\text{ atm}\end{array}$

And,

$\begin{array}{l}{V}_A=2\text{ litres}\\ V_B=9\text{ litres}\end{array}$

Here, $P_A$ pressure at the point A, $V_A$ represent the volume at point A, $P_B$ represents the pressure at the point B, and $V_B$ represent the volume at point B.

Recall the gas equation between points A and B.

$\frac{P_A{V}_A}{T_A}=\frac{P_B{V}_B}{T_B}$

Here, $T_B$ is the temperature at point B in Kelvin.

Substitute $4\text{ atm}$ for $P_B$, $9\text{ litres}$ for $V_B$, $6\text{ atm}$ for $P_A$, $473\text{ K}$ for $T_A$, and $2\text{ litres}$ for $V_A$

$\begin{array}{c}\frac{\left(6\text{ atm}\right)\left(2\text{ litres}\right)}{473\text{ K}}=\frac{\left(4\text{ atm}\right)\left(9\text{ litres}\right)}{T_B}\\ T_B=\frac{\left(4\text{ atm}\right)\left(9\text{ litres}\right)}{\left(6\text{ atm}\right)\left(2\text{ litres}\right)}\left(473\text{ K}\right)\\ \simeq 1.42\times {10}^3\text{ K}\end{array}$

Conclusion:

The temperature at point B is $1.42\times {10}^3\text{ K}$.

To determine

The value of $\Delta U$ for the $25.0\text{ g}$ of gas enclosed in the cylinder for the portion from A to B shown in Fig. 20-6.

Answer to Problem 52SP

Solution:

$14.9\text{ kJ}$

Explanation of Solution

Given Data:

Refer to Fig. 20-6.

The temperature at A is $200°\text{C}$.

The mass of the ideal gas enclosed is $25.0\text{ g}$.

The specific heat of the gas at constant volume is $0.150\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$

Formula Used:

The formula for change in internal energy is,

$\Delta U=m{c}_v\Delta T$

Here, $\Delta U$ is the change in internal energy during the process, $\Delta T$ is the change in temperature during the process, $c_v$ is the specific heat at constant volume, and $m$ is the mass of the body.

The formula for conversion of temperature of gas from Kelvin scale to Celsius scale is,

$t=\left[T-273\right]°\text{C}$

Here, $t$ is the temperature in Kelvin and $T$ is the temperature in degree Celsius.

Explanation:

Draw the thermodynamic cycle diagram given in Fig- 20.6.

Calculate the temperature at point B in Celsius.

$t_B=\left[T_B-273\right]°\text{C}$

Here, $t_B$ is the temperature at point B in Celsius.

Substitute $1.42\times {10}^3\text{ K}$ for $T_B$.

$\begin{array}{c}{t}_B=\left[1.42\times {10}^3\text{ K}-273\right]\text{°C}\\ =1147°\text{C}\end{array}$

Calculate the change in temperature from A to B.

$\Delta T=t_B-t_A$

Substitute $1147°\text{C}$ for $t_B$ and $200°\text{C}$ for $t_A$.

$\begin{array}{c}\Delta T=1147°\text{C}-200°\text{C}\\ =9\text{47}°\text{C}\end{array}$

Recall the formula for change in internal energy.

$\Delta U=m{c}_v\Delta T$

Substitute $0.150\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$ for $c_v$, $9\text{47}°\text{C}$ for $\Delta T$, and $25.0\text{ g}$ for $m$.

$\begin{array}{c}\Delta U=\left(25.0\text{ g}\right)\left[0.150\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(9\text{47}°\text{C}\right)\\ =3551.25\text{ cal}\\ =3551.25\text{ cal}\left(\frac{4.186\text{ J}}{1\text{ cal}}\right)\left(\frac{0.001\text{ kJ}}{1\text{ J}}\right)\\ \simeq 14.9\text{ kJ}\end{array}$

Conclusion:

The value of $\Delta U$ for the gas from A to B is $14.9\text{ kJ}$.

To determine

The value of $\Delta W$ for the $25.0\text{ g}$ of gas enclosed in the cylinder for the portion from A to B shown in Fig. 20-6.

Answer to Problem 52SP

Solution:

$3.55\text{ kJ}$

Explanation of Solution

Given Data:

Refer to Fig. 20-6.

The temperature at A is $200°\text{C}$.

The mass of the ideal gas enclosed is $25.0\text{ g}$.

The specific heat of the gas at constant volume is $0.150\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$

Formula Used:

The area of a trapezium is calculated by the formula,

$A=\frac{1}{2}\left(a+b\right)\left(h\right)$

Here, $A$ is the area of the trapezium, $a$ and $b$ are the lengths of the parallel sides of the trapezium, $a+b$ represents the sum of the parallel sides, and $h$ is the perpendicular distance between the parallel sides of the trapezium.

The work done in a thermodynamic process is given by the area under the line representing the process in the pressure-volume diagram.

$\Delta W=A_{PVD}$

Here, $A_{PVD}$ is the area under the line representing the process in the pressure-volume diagram and $\Delta W$ is the work done in the process.

Explanation:

Draw the thermodynamic cycle diagram given in Fig- 20.6.

Understand that the work done in the thermodynamic process AB is equal to the area of the pressure-volume diagram under the line AB.

Draw the thermodynamic cycle diagram showing the area under the line AB.

Here, the points E and F are as shown in the figure and the work done during the process AB is represented by the area under the line AB, which is equal to the area of the trapezium ABEF.

Refer to the figure and write the values of the lengths of sides AF, BE, and EF.

$\begin{array}{c}AF=6\text{ atm}-0\text{ atm}\\ =\text{6 atm}\end{array}$

$\begin{array}{c}BE=4\text{ atm}-0\text{ atm}\\ =4\text{ atm}\end{array}$

$\begin{array}{c}EF=\text{9 litres}-2\text{ litres}\\ =7\text{ litres}\end{array}$

Recall the expression for the area of trapezium ABEF to calculate the area under the line AB in order to calculate the work done in the process AB.

$A_1=\frac{1}{2}\left(AF+BE\right)\left(EF\right)$

Here, $A_1$ is the area of the trapezium ABEF.

Substitute $\text{6 atm}$ for $AF$, $\text{4 atm}$ for $BE$, and $7\text{ litres}$ for $EF$.

$\begin{array}{c}{A}_1=\frac{1}{2}\left(\text{6 atm}+\text{4 atm}\right)\left(7\text{ litres}\right)\\ =\frac{1}{2}\left(10\text{ atm}\right)\left(7\text{ litres}\right)\\ =\frac{1}{2}\left[\left(10\text{ atm}\right)\left(\frac{1.013\times {10}^5\text{ Pa}}{1\text{ atm}}\right)\right]\left[\left(7\text{ litres}\right)\left(\frac{{10}^{-3}{\text{ m}}^3}{1\text{ litre}}\right)\right]\\ =3545.5\text{ J}\end{array}$

Recall the expression for the net-work done in a thermodynamic process in terms of the area of the pressure-volume diagram.

$\Delta W=A_1$

Substitute $3545.5\text{ J}$ for $A_1$.

$\begin{array}{c}\Delta W=3545.5\text{ J}\\ =3545.5\text{ J}\left(\frac{0.001\text{ kJ}}{1\text{ J}}\right)\\ \simeq 3.55\text{ kJ}\end{array}$

The work done from A to B is $3.55\text{ kJ}$.

Conclusion:

The value of $\Delta W$ for the gas from A to B is $3.55\text{ kJ}$.

To determine

The value of $\Delta Q$ for the $25.0\text{ g}$ of gas enclosed in the cylinder for the portion from A to B shown in Fig. 20-6.

Answer to Problem 52SP

Solution:

$18.4\text{ kJ}$

Explanation of Solution

Given Data:

Refer to Fig. 20-6.

The temperature at A is $200°\text{C}$.

The mass of the ideal gas enclosed is $25.0\text{ g}$.

The specific heat of the gas at constant volume is $0.150\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$

Formula Used:

The first law of thermodynamics for a process is written as,

$\Delta Q=\Delta U+\Delta W$

Here, $\Delta W$ is the work done by the system during the process, $\Delta U$ is the change in internal energy during the process and $\Delta Q$ is the heat supplied to the system during the process.

Explanation:

Draw the thermodynamic cycle diagram given in Fig- 20.6.

Recall the expression for the first law of thermodynamics for the process AB.

$\Delta Q=\Delta U+\Delta W$

Substitute $14.9\text{ kJ}$ for $\Delta U$ and $3.55\text{ kJ}$ for $\Delta W$.

$\begin{array}{c}\Delta Q=14.9\text{ kJ}+3.55\text{ kJ}\\ \simeq 18.4\text{ kJ}\end{array}$

Conclusion:

The value of $\Delta Q$ for the gas from A to B is $18.4\text{ kJ}$.