# The gas inside of a container exerts 8 Pa of pressure and is at a temperature of 220 ^o K. If the temperature of the gas changes to 20 ^oC with no change in the container's volume, what is the new pressure of the gas?

May 25, 2016

The final pressure is $10.66$ Pa.

#### Explanation:

I assume that we are working with a perfect gas.
The process is called isochoric (it means that mantains the volume unchanged). Those process follow the law $\frac{P}{T} =$constant.

This mean that if we have an initial pressure and temperature ${P}_{i} , {T}_{i}$ and a final pressure and temperature ${P}_{f} , {T}_{f}$ the law tell us:
${P}_{i} / {T}_{i} =$constant
${P}_{f} / {T}_{f} =$constant
then
${P}_{i} / {T}_{i} = {P}_{f} / {T}_{f}$.
In our case we have
${P}_{i} = 8$ Pa
${T}_{i} = 220$ K (you shouldn't use the symbol $\setminus \circ$ for Kelvin)
${T}_{f} = {20}^{\setminus} \circ$ C
and we want to know ${P}_{f}$.

It is always better to work in the International System then we convert the Celsius in Kelvin adding $273.15$: ${T}_{f} = {20}^{\setminus} \circ$ C $= 293.15$ K.

Before solving the equation of the isochoric process we can notice that the temperature is increasing of 73.15 K from 220 K to 293.15 K. If we warm up a gas we expect that the pressure increase because the particles of the gas move faster. So at the end we should find something bigger than 8 Pa.

We can write our equation for isochoric process
$\frac{8}{220} = {P}_{f} / 293.15$ obtaining
${P}_{f} = \frac{8}{220} 293.15 = 10.66$ Pa (that is bigger than 8, as expected from a warmer gas).