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?

1 Answer
May 25, 2016

Answer:

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).