Question

From a sample of an ideal gas at 15.0 atm and 10.0 L is allowed to expand against a constant external pressure of 2.00 atm at a constant temperature. Calculate the work in units of kJ for the gas expansion. (Hint: Boyle’s law applies.)

Short answer

The workfor the gas expansion from a sample of an ideal gas at 15.0atm and10.0L that is allowed to expand against a constant external pressureof 2.00 atm at a constanttemperatureis$-\text{13.2}\text{kJ}$

Step-by-step solution

Thedetermination of the Final Volume.

By using the equation of Boyle Law,

$\begin{array}{rcl}& & {\text{P}}_{\text{1}}{\text{V}}_{\text{1}}\text{=}{\text{P}}_{\text{2}}\text{V}\\ & & {\text{(15.0 atm×10.0 L) = (2.00 atm×V}}_{\text{2}}\text{)}\end{array}$

Therefore, the final volume becomes

$$\begin{gathered} {\text{V}}_{\text{2}}\text{=}\frac{{\displaystyle \text{(15.0 atm×10.0 L)}}}{{\displaystyle \text{2.00 atm}}} \\ \text{= 75.0}\text{L} \end{gathered}$$

Thedetermination of Change in Volume.

$$\begin{gathered} ∆\text{V}\text{=}{\text{V}}_{\text{2}}-{\text{V}}_{\text{1}} \\ \text{=}\text{75.0}\text{L}-\text{10.0}\text{L} \\ \text{=}\text{65.0}\text{L} \end{gathered}$$

Thedetermination of the Work for the Gas Expansion.

As the temperature is constant the gas performs work on the surroundings.

The work of the gas expansion is therefore,\

$$\begin{gathered} \text{w}\text{=}-\text{P∆V} \\ \text{=}-\text{(2.00}\text{atm}\times \text{65.0}\text{L)} \\ \text{=}-\text{130}\text{L - atm} \end{gathered}$$

Step 4:The conversation of Units to kilo joules (kJ)

As,$\text{w}\text{=}-\text{130}\text{L - atm}$

$$\begin{gathered} \text{w (joules) =}-\text{130 L -}\text{atm}\times \frac{\text{101.33} \text{J}}{\text{1 L - atm}} \\ \text{=}-\text{13169}\text{J} \\ \text{=}-\text{13.2}\text{kJ} \end{gathered}$$

Hence, the work in units of kJ for the gas expansion is $-\text{13.2}\text{kJ}$.