Chapter 10, Problem 6SP

Suppose the pressure of an ideal gas mixture remains constant at 1800 Pa (1 Pa = 1 N/m²) and the temperature is increased from 220K to 1100K.

a.    If the original volume of the gas was 0.17 m³, what is the final volume? (See example box 10.3.)

b.    What is the change in volume ΔV for this process?

c.    How much work does the gas do on the surroundings during the expansion?

d.    If the initial volume was 0.24m³ and the same temperature change occurred, would the work done be the same as in the first case? Show by repeating the steps of the first three parts.

e.    Is the same amount of gas involved in these two situations? Explain.

To determine

The final volume.

Answer to Problem 6SP

The final volume is $0.85{\text{m}}^3$.

Explanation of Solution

Given info: The pressure of an ideal gas mixture is $1800\text{Pa}$ and the temperature increased from $220\text{K}$ to $1100\text{K}$.

Write the equation satisfied by idea gas at two different pressure, volume and temperature

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$ (1)

Here,

$P_1$ is the initials pressure

$P_2$ is the final pressure

$V_1$ is the initial volume

$V_2$ is the final volume

$T_1$ is the initial temperature

$T_2$ is the final temperature

The pressure is remains constant hence, $P_1=P_2$ thus, equation (1) will be rewritten as

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$ (2)

Rearrange equation (2) to obtain an expression for final volume

$V_2=\frac{V_1{T}_2}{T_1}$ (3)

Substitute $0.17{\text{m}}^3$ for $V_1$, $220\text{K}$ for $T_1$ and $1100\text{K}$ for $T_2$ in equation (3)

$\begin{array}{c}{V}_2=\frac{0.17{\text{m}}^3\times 1100\text{K}}{220\text{K}}\\ =0.85{\text{m}}^3\end{array}$

Conclusion:

The final volume is $0.85{\text{m}}^3$.

To determine

The change in the volume for the process.

Answer to Problem 6SP

The change in the volume for the process is $0.68{\text{m}}^3$.

Explanation of Solution

Given info:

Write the expression for the change in volume

$\Delta V=V_2-V_1$

Substitute $0.17{\text{m}}^3$ for $V_1$ and $0.85{\text{m}}^3$ for $V_2$ in the above equation

$\begin{array}{c}\Delta V=0.85{\text{m}}^3-0.17{\text{m}}^3\\ =0.68{\text{m}}^3\end{array}$

Conclusion:

The change in the volume for the process is $0.68{\text{m}}^3$.

To determine

The work done by the gas on the surroundings during the expansion.

Answer to Problem 6SP

The work done by the gas on the surroundings during the expansion is $1224\text{J}$.

Explanation of Solution

Given info:

Write the expression for work done in terms of volume and temperature

$W=P\Delta V$

Here,

$W$ is the work done

$\Delta V$ is the change in volume

Substitute $1800\text{Pa}$ for $P$ and $0.68\text{}{\text{m}}^3$ for $\Delta V$ in the above equation

$\begin{array}{c}W=1800\text{Pa}\times 0.68\text{}{\text{m}}^3\\ =1224\text{J}\end{array}$

Conclusion:

The work done by the gas on the surroundings during the expansion is $1224\text{J}$.

To determine

The work done if the initial temperature is $0.24{\text{m}}^3$.

Answer to Problem 6SP

The final volume will be $0.96{\text{m}}^3$, if the initial volume is $0.24{\text{m}}^3$.

Explanation of Solution

Given info: The initial volume is $0.24{\text{m}}^3$.

Write the expression for final volume

$V_2=\frac{V_1{T}_2}{T_1}$ (1)

Substitute $0.24{\text{m}}^3$ for $V_1$, $220\text{K}$ for $T_1$ and $1100\text{K}$ for $T_2$ in the above equation

$\begin{array}{c}{V}_2=\frac{0.24{\text{m}}^3\times 1100\text{K}}{220\text{K}}\\ =1.2{\text{m}}^3\end{array}$

Write the expression for the change in volume

$\Delta V=V_2-V_1$ (2)

Substitute $0.24{\text{m}}^3$ for $V_1$ and $1.2{\text{m}}^3$ for $V_2$ in the above equation

$\begin{array}{c}\Delta V=1.2{\text{m}}^3-0.24{\text{m}}^3\\ =0.96{\text{m}}^3\end{array}$

Conclusion:

The final volume will be $0.96{\text{m}}^3$, if the initial volume is $0.24{\text{m}}^3$.

To determine

To explain is the same amount of gas involved in these two situations.

Answer to Problem 6SP

The amount of gas involved in the two situations will be different.

Explanation of Solution

Given info:

Write the expression for ideal gas equation

$PV=NkT$

Here,

$P$ is the pressure

$V$ is the volume

$N$ is the number of molecules

$k$ is the Boltzmann Constant

In this case, both pressure and temperature is remains as same, but there is a change in volume. According to the above equation, the number of molecules will be different for different values of $V$ at constant pressure and temperature.

Conclusion:

Different amount of gas will be involved in both case since, the volume is changing at constant pressure and temperature.