Question

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`?

Answer 1

The pressure will fall to `p_2~~0.523` `atm`

Explanation:

According to the ideal gas equation we know that:

`(pV)/T=const`

so if we write that:

`p_1=1atm`

`v_1=85cm^3`

`T_1=310K`

After the change we have:

`v_2=180cm^3`

`T_2=280K`

We have to calculate the new pressure `p_2`

Using the equation stated at the beginning we have:

`(p_1*v_1)/T_1=(p_2*v_2)/T_2`

If we transform the formula we get:

`p_2=(p_1*v_1*T_2)/(T_1*v_2)`

If we substitute the given data we get:

`p_2=(1*85*310)/(180*280)=2635/5040~~0.523`