Question

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?

Answer 1

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 `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 `\circ` for Kelvin)
`T_f=20^\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^\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
`8/220 = P_f/293.15` obtaining
`P_f=8/220 293.15=10.66` Pa (that is bigger than 8, as expected from a warmer gas).