Question

A sample of ideal gas expands from an initial pressure and volume of 32atmand$\mathbf{1}\mathbf{.}\mathbf{0}\mathit{L}$to a final volume of$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{L}$. The initial temperature is$\mathbf{300}\mathbf{K}$. If the gas is monatomic and the expansion isothermal, what are the (a) final pressure${\mathbf{P}}_f$, (b) final temperature${\mathbf{T}}_f$, and (c) work W done by the gas? If the gas is monatomic and the expansion adiabatic, what are (d)${\mathbf{P}}_f$, (e)${\mathbf{T}}_f$, and (f) W? If the gas is diatomic and the expansion adiabatic, what are (g)${\mathbf{P}}_f$, (h)${\mathbf{T}}_f$, and (i) W?

Short answer

1. For the expansion of monoatomic gas, isothermally,the final pressure is${\text{P}}_f=8.0\text{\hspace{0.17em}atm}$.
2. For the expansion of monoatomic gas, isothermally,the final temperature is${\text{T}}_f=300\text{\hspace{0.17em}K}$.
3. For the expansion of monoatomic gas, isothermally,the work done by the gas is.$\text{W}=4.5\text{\hspace{0.17em}kJ}$
4. Foradiabatic expansionof monoatomic gas, the final pressureis.${\text{P}}_f=3.15\text{atm}$
5. Foradiabatic expansionof monoatomic gas, the final temperature is.${\text{T}}_f=118\text{\hspace{0.17em}K}$
6. Foradiabatic expansionof monoatomic gas,the work done by the gas is.$\text{W}=2.9\text{kJ}$
7. Foradiabatic expansionof diatomic gas, the final pressureis.${\text{P}}_f=4.6\text{\hspace{0.17em}atm}$
8. Foradiabatic expansionof diatomic gas, the final temperature is.${\text{T}}_f=172.3\text{\hspace{0.17em}K}$
9. For adiabatic expansionof diatomic gas, the work done by the gas is.$\text{W}=3.44\text{kJ}$

Step-by-step solution

Concept

The process conducted at constant temperature is called the isothermal process. Practically such processes are conducted very slowly so that sufficient time is available for heat exchange. In isothermal process the state parameters must satisfy the relation-

${\mathbf{P}}_1{\mathbf{V}}_1\mathbf{=}{\mathbf{P}}_2{\mathbf{V}}_2\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Here,$\mathbf{P}\mathbf{and}\mathbf{V}$ are pressure and volume of a gas and the subscripts$\mathbf{1}\mathbf{and}\mathbf{2}$ represents the initial and final value of the parameter respectively.

Whereas, the process that is conducted very rapidly is an adiabatic process. In this process no exchange of heat takes place between system and surrounding. The state parameters must satisfy the relation-

$\begin{array}{l}{\mathbf{P}}_1{\mathbf{V}}_1^{\gamma }\mathbf{=}{\mathbf{P}}_2{\mathbf{V}}_2^{\gamma }\text{\hspace{0.17em}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\\ {\mathbf{T}}_i{\mathbf{V}}_i^{\gamma \mathbf{-}\mathbf{1}}\mathbf{=}{\mathbf{T}}_f{\mathbf{V}}_f^{\gamma \mathbf{-}\mathbf{1}}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)\end{array}$

Given Data

1. The initial pressure of the ideal gas:${\text{p}}_i=32\text{\hspace{0.17em}atm}$
2. The initial temperature of the ideal gas:${\text{T}}_i=300\text{K}$
3. The initial volume of the ideal gas:${\text{V}}_i=1.0\text{\hspace{0.17em}L}=1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^3$
4. The final value of volume of an ideal gas:${\text{V}}_f=4.0\text{\hspace{0.17em}L}=4.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^3$
5. The ideal gas constant:$R=8.31\text{\hspace{0.17em}J/mol.K}$

Calculations

When the monatomic gas expands isothermally,

(a)

Using equation (1), we have-

$$\begin{gathered} {\text{p}}_i{\text{V}}_i={\text{p}}_f{\text{V}}_f \\ {\text{p}}_f=\frac{{\text{p}}_i{\text{V}}_i}{{\text{V}}_f} \end{gathered}$$

For the given values of variables, the above equation becomes-

$\begin{array}{c}{p}_f=\frac{(32\text{\hspace{0.17em}atm})\times (1.01\times {10}^5\text{\hspace{0.17em}Pa}/\text{atm})\times (1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}{(4.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}\\ =8.08\times {10}^5\text{\hspace{0.17em}Pa}\\ =8\text{\hspace{0.17em}atm}\end{array}$

(b)

As the process is isothermal, the temperature of the gas remains constant. Hence,

$\begin{array}{c}{\text{T}}_i={\text{T}}_f\\ =300\text{\hspace{0.17em}K}\end{array}$

(c)

Next, we need to determine the number of moles first. For this, we use the ideal gas equation as

$\begin{array}{l}pV=nRT\\ n=\frac{pV}{RT}\end{array}$

For the given values of variables, the above equation becomes-

$\begin{array}{c}n=\frac{(32\text{\hspace{0.17em}atm})\times (1.01\times {10}^5\text{\hspace{0.17em}Pa}/\text{atm})\times (1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}{(8.31\text{\hspace{0.17em}J/mol.K})\times \left(300\text{\hspace{0.17em}K}\right)}\\ =1.3\text{\hspace{0.17em}mol}\end{array}$

Thework done in isothermal expansion is

$W=nRT\ln \frac{V_f}{V_i}$

For the given values, we have-

$\begin{array}{c}W=\left(1.3\text{\hspace{0.17em}mol}\right)\times (8.31\text{\hspace{0.17em}J/mol.K})\times \left(300\text{\hspace{0.17em}K}\right)ln\left(\frac{4.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}}}{1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}}}\right)\\ =4.5\text{\hspace{0.17em}kJ}\end{array}$

For adiabatic expansion of monatomic ideal gas-

(d)

Using equation (2)-

$\begin{array}{l}{p}_i{V}_i^{\gamma }=p_f{V}_f^{\gamma }\\ p_f=\frac{p_i{V}_i^{\gamma }}{V_f^{\gamma }}\end{array}$

Here,$\gamma =1.67$

So,for the given values of variables, the above equation becomes-

$\begin{array}{c}{p}_f=\frac{(32\text{\hspace{0.17em}atm})\times (1.01\times {10}^5\text{\hspace{0.17em}Pa}/\text{atm})\times {(1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{1.67}}{{(4.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{1.67}}\\ =3.19\times {10}^5\text{Pa}\\ =3.15\text{\hspace{0.17em}atm}\end{array}$

(e)

Using equation (3)-

$\begin{array}{l}{T}_i{V}_i^{\gamma -1}=T_f{V}_f^{\gamma -1}\\ T_f=\frac{T_i{V}_i^{\gamma -1}}{V_f^{\gamma -1}}\end{array}$

For the given values of variables, the above equation becomes-

$\begin{array}{c}{T}_f=\frac{(300\text{\hspace{0.17em}K})\times {(1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{1.67-1}}{{(4.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{1.67-1}}\\ =118\text{\hspace{0.17em}K}\end{array}$

(f)

The work done by the gas in adiabatic expansion is calculated as

$W=\frac{1}{-\gamma +1}{p}_i{V}_i^{\gamma }[V_f^{-\gamma +1}-V_i^{-\gamma +1}]$

For the given values of variables, the above equation becomes-

$$\begin{gathered} W=\left(\begin{array}{l}\frac{(32\text{\hspace{0.17em}atm})\times (1.01\times {10}^5\text{\hspace{0.17em}Pa}/\text{atm})\times {(1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{1.67}}{-1.67+1}\times \\ [{(4.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{-1.67+1}-{(1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{-1.67+1}]\end{array}\right) \\ =2.9\times {10}^3\text{\hspace{0.17em}J} \end{gathered}$$

For adiabatic expansion of diatomic ideal gas

(g)

Using equation (2),

$\begin{array}{l}{p}_i{V}_i^{\gamma }=p_f{V}_f^{\gamma }\\ p_f=\frac{p_i{V}_i^{\gamma }}{V_f^{\gamma }}\end{array}$

Here,$\gamma =1.4$

So,for the given values of variables, the above equation becomes-

$\begin{array}{c}{p}_f=\frac{(32\text{\hspace{0.17em}atm})\times (1.01\times {10}^5\text{\hspace{0.17em}Pa}/\text{atm})\times {(1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{1.4}}{{(4.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{1.4}}\\ =4.6\times {10}^5\text{Pa}\\ =4.6\text{\hspace{0.17em}atm}\end{array}$

(h)

Using equation (3)-

$\begin{array}{l}{T}_i{V}_i^{\gamma -1}=T_f{V}_f^{\gamma -1}\\ T_f=\frac{T_i{V}_i^{\gamma -1}}{V_f^{\gamma -1}}\end{array}$

For the given values of variables, the above equation becomes-

$\begin{array}{c}{T}_f=\frac{(300\text{\hspace{0.17em}K})\times {(1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{1.4-1}}{{(4.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{1.4-1}}\\ =172.3\text{\hspace{0.17em}K}\end{array}$

(i)

The work done by the gas in adiabatic expansion is calculated as-

$W=\frac{1}{-\gamma +1}{p}_i{V}_i^{\gamma }[V_f^{-\gamma +1}-V_i^{-\gamma +1}]$

For the given values of variables, the above equation becomes-

$$\begin{gathered} W=\left(\begin{array}{l}\frac{(32\text{\hspace{0.17em}atm})\times (1.01\times {10}^5\text{\hspace{0.17em}Pa}/\text{atm})\times {(1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{1.4}}{-1.4+1}\times \\ [{(4.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{-1.4+1}-{(1.0\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}})}^{-1.4+1}]\end{array}\right) \\ =3.44\times {10}^3\text{\hspace{0.17em}J} \end{gathered}$$

Conclusion

The values of final pressure $\left(p_f\right)$, final temperature $\left(T_f\right)$and the work done$\left(W\right)$ for the three cases are-

• For the expansion of monoatomic gas, isothermally- $p_f=8\text{\hspace{0.17em}atm},{\text{\hspace{0.17em}T}}_f=300\text{\hspace{0.17em}K},\text{\hspace{0.17em}and\hspace{0.17em}}$.$W=4.5\text{\hspace{0.17em}kJ}$
• For adiabatic expansion of monoatomic ideal gas-.$p_f=3.15\text{\hspace{0.17em}atm},{\text{\hspace{0.17em}T}}_f=118\text{\hspace{0.17em}K},\text{\hspace{0.17em}and\hspace{0.17em}}$$W=2.9\times {10}^3\text{\hspace{0.17em}J}$
• For adiabatic expansion of diatomic ideal gas-.$p_f=4.6\text{\hspace{0.17em}atm},{\text{\hspace{0.17em}T}}_f=172.3\text{\hspace{0.17em}K},\text{\hspace{0.17em}and\hspace{0.17em}}$$W=3.44\times {10}^3\text{\hspace{0.17em}J}$