Question

One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are removed suddenly from the piston to give the following sequence of three pressures:

1. P₁ = 5.00 atm (initial state)
2. b. P₂ = 2.24 atm
3. P₃ = 1.00 atm (final state)

What is the total work (in joules) in going from the initial to the final state by way of the preceding two steps? What would be the total work if the process were carried out reversibly?

Short answer

The total work (in joules) in going from the initial to the final state by way of thepreceding two steps is$\mathbf{-}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{21}\mathbf{}\mathit{k}\mathit{J}$.

Thetotal work if the process were carried out reversiblyis $\mathbf{-}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{68}\mathbf{}\mathit{k}\mathit{J}$.

Step-by-step solution

Step 1: Introduction to the Concept

The mathematical formula for calculating work done,

$\mathit{w}\mathbf{=}\mathbf{-}{\mathit{P}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathit{\Delta }\mathit{V}$

Where w = work done, P = pressure, and $\Delta V=$Volume change

In a reversible process, the following expression is used to compute work:

${\mathit{w}}_{\mathbf{r}\mathbf{e}\mathbf{v}}\mathbf{=}\mathbf{-}\mathit{n}\mathit{R}\mathit{T}\mathit{l}\mathit{n}\frac{{\mathbf{V}}_{\mathbf{f}}}{{\mathbf{V}}_{\mathbf{t}}}$

$\mathit{P}\mathit{V}\mathbf{=}\mathit{n}\mathit{R}\mathit{T}$is the ideal gas law.

Where, P = pressure, V = volume, n = number of moles, R = Universal gas constant, and T = temperature.

Step 2: Determination of work done for this two step pressure

In a cylinder with a moveable piston, one mole of an ideal gas is present. The piston's weights are removed to provide three pressure measurements.

From the given,

Number of moles$=1.0\text{mole}$

Temperature $=77°C$

P₁ = 5.00 atm (initial state)
P₂ = 2.24 atm
P₃ = 1.00 atm (final state)

Here the ideal gas law in terms of Volume,

$\mathit{V}\mathbf{=}\frac{\mathbf{n}\mathbf{R}\mathbf{T}}{\mathbf{P}}$

Let’s convert the temperature from degree Celsius to Kelvin

So, $T=350\text{K}$

Let’s substitute all values,

$V_1=\frac{1.00\text{moles}\times \text{0.8206L atm/Kmoles}\times 350\text{K}}{5.00\text{atm}}$

$V_1=5.74\text{L}$

$V_2=\frac{1.00\text{moles}\times \text{0.8206Latm/Kmoles}\times 350\text{K}}{2.24\text{atm}}$

$V_2=12.8\text{L}$

$V_3=\frac{1.00\text{moles}\times \text{0.8206Latm/K}mole\times 350\text{K}}{1.00\text{atm}}$

$V_3=28.7\text{L}$

Thus the work done for two step pressure is,

$\mathit{w}\mathbf{=}\mathbf{-}{\mathit{P}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathit{\Delta }\mathit{V}$

Step 3: Determination of total work done by irreversible process

1st change:

$w_{1\to \text{2}}=-P_{ext}\left(V_2-V_1\right)$

Let us substitute all values,

$w_{1\to \text{2}}=-\left(2.24\text{atm}\right)\left(12.8\text{L}-5.74\text{L}\right)$

$=-1.60\text{KJ}$

2^nd change:

$w_{2\to \text{3}}=-P_{ext}\left(V_3-V_2\right)$

Let us substitute all values,

$w_{2\to \text{3}}=-\left(1.00\text{atm}\right)\left(28.7\text{L}-12.8\text{L}\right)$

$=-1.61\text{KJ}$

Work done for two processes,

$$\begin{gathered} w=w_{1\to 2}+w_{2\to 3} \\ =-1.60 KJ-1.61 KJ \\ =-3.21\text{KJ} \end{gathered}$$

Therefore, the total work done by the two processes are $\mathbf{-}\mathbf{}\mathbf{3}\mathbf{.}\mathbf{21}\mathbf{}\mathit{k}\mathit{J}$.

Step 4: Determination of total work done by reversible process

For reversible work process, ${\mathit{w}}_{\mathbf{r}\mathbf{e}\mathbf{v}}\mathbf{=}\mathbf{-}\mathit{n}\mathit{R}\mathit{T}\mathit{l}\mathit{n}\frac{{\mathbf{V}}_{\mathbf{f}}}{{\mathbf{V}}_{\mathbf{t}}}$

$w_{rev}=-\left(1.00\text{moles}\right)\times (8.3145J/Kmol)\times \left(350\text{K}\right)\times \left(\ln \frac{28.7\text{L}}{5.74\text{L}}\right)$

$=-4683.474\text{J}$

Conversion from Joule to kilojoules,

$w_{rev}=-4683.474\text{J}\times \frac{1\text{KJ}}{1000\text{J}}$

$=-4.68\text{kJ}$

Therefore the work for reversible process is$\mathbf{-}\mathbf{}\mathbf{4}\mathbf{.}\mathbf{68}\mathbf{}\mathit{k}\mathit{J}$