# A canister containing air has a volume of 85 cm^3 and a pressure of 1.45 atm when the temperature is 310 K. What is the pressure when the volume is increased to 180 cm^3 and the temperature is reduced to 280 K?

Mar 2, 2017

The pressure will fall to ${p}_{2} \approx 0.523$ $a t m$

#### Explanation:

According to the ideal gas equation we know that:

## $\frac{p V}{T} = c o n s t$

so if we write that:

${p}_{1} = 1 a t m$

${v}_{1} = 85 c {m}^{3}$

${T}_{1} = 310 K$

After the change we have:

${v}_{2} = 180 c {m}^{3}$

${T}_{2} = 280 K$

We have to calculate the new pressure ${p}_{2}$

Using the equation stated at the beginning we have:

$\frac{{p}_{1} \cdot {v}_{1}}{T} _ 1 = \frac{{p}_{2} \cdot {v}_{2}}{T} _ 2$

If we transform the formula we get:

${p}_{2} = \frac{{p}_{1} \cdot {v}_{1} \cdot {T}_{2}}{{T}_{1} \cdot {v}_{2}}$

If we substitute the given data we get:

${p}_{2} = \frac{1 \cdot 85 \cdot 310}{180 \cdot 280} = \frac{2635}{5040} \approx 0.523$