Question

A gas that has a volume of 13 liters, a temperature of 25 0C, and an unknown pressure has its volume increased to 27 liters and its temperature decreased to 15 0C. If I measure the pressure after the change to be 1.3 atm, what was the original pressure of the gas?

Answer 1

This is an exercise in the general or combined gas law.

To start solving this exercise, we obtain the data:

Data:

• V₁ = 13 Lt
• T₁ = 25 °C + 273 = 298 k
• V₂ = 27 Lt
• T₂ = 15 °C + 273 = 288 k
• P₁ = 1.3 atm
• P₂ = ¿?

We use the following formula:

• P₁V₁T₂ = P₂V₂T₁ ⇒ General Formula

Where

• P₁ = Initial pressure
• V₁ = Initial volume
• T₂ = Initial temperature
• P₂ = Final pressure
• V₂ = Final volume
• T₁ = Initial temperature

We clear the general formula for the final pressure.

`\large\displaystyle\text{$\begin{gathered}\sf P_(2)=(P_(1)V_(1)T_(2) )/(V_(2)T_(1)) \ \to \ Clear \ formula \end{gathered}$}`

We substitute our data into the formula to solve:

`\large\displaystyle\text{$\begin{gathered}\sf P_(2)=\frac{(1.3 \ atm)(13\\ot{l})(288\\ot{k} )}{(27 \\ot{l})(298 \\ot{K})} \end{gathered}$}`

`\large\displaystyle\text{$\begin{gathered}\sf P_(2)=(4867.2)/(8046) \ atm \end{gathered}$}`

`\boxed{\large\displaystyle\text{$\begin{gathered}\sf P_(2)=0.604 \ atm \end{gathered}$}}`

If I measure the pressure after the change by 1.3 atm, the original pressure of the gas will be 0.604 atm.