Question

Consider a mixture of air and gasoline vapor in a cylinder with a piston. The original volume is$\mathbf{40}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{3}}$. If the combustion of this mixture releases 950. J of energy, to what volume will the gases expand against a constant pressure of 650.torr if all the energy of combustion is converted into work to push back the piston?

Short answer

The volume of the gases expands against a constant pressure of 650 torrs. if all the energy of combustion is converted into work to push back the piston is $1.162\text{L}$.

Step-by-step solution

Conversion of Units.

In this step, we will convert the units of volumeand also units for pressure.

Volume conversion

$$\begin{gathered} {\text{1cm}}^3=0.001\text{L} \\ {\text{40cm}}^3=0.04\text{L} \end{gathered}$$

Pressure conversion

$$\begin{gathered} \text{1atm}\text{=}\frac{1\text{torr}}{7\text{60}\text{mm}\text{Hg}} \\ 650\text{torr}=\frac{650\text{torr}}{7\text{60}\text{mm}\text{Hg}}=0.855\text{atm} \end{gathered}$$

Calculation of Volume.

The amount of work done if all the energy of combustion is converted into work to push back the piston is

Given,$\text{W}=960\text{J}\text{=}\text{0.96kJ}$

$$\begin{gathered} \text{W}\text{=}\text{P∆V} \\ {\text{= P(V}}_{\text{2}}{\text{-V}}_{\text{1}}\text{)} \end{gathered}$$

After putting the values, we get

$\text{0.96 kJ}\text{=}{\text{0.855 atm (V}}_{\text{2}}\text{- 0.04 L)}$

The volume found is

$$\begin{gathered} {\text{(V}}_{\text{2}}\text{- 0.04 L)}\text{=}\frac{\text{0.96}}{\text{0.855}} \\ {\text{(V}}_{\text{2}}\text{- 0.04 L)}\text{=}\text{1.122} \\ {\text{V}}_{\text{2}}\text{= 1.122}\text{+}\text{0.04} \\ {\text{V}}_{\text{2}}\text{= 1.162}\text{L} \end{gathered}$$

Therefore, $1.162\text{L}$is the volume of the gases that expands against a constant pressure of 650 torrs. if all the energy of combustion is converted into work to push back the piston.