Question

A `73*mL` volume of gas enclosed in a piston at a pressure of `100*kPa`, and a temperature of `0` `""^@C` was heated to a temperature of `80` `""^@C` and the new volume occupied was `4530*mL`. What is the new pressure that the gas exerts?

Answer 1

Well for a given molar quantity of gas, `(P_1V_1)/T_1=(P_2V_2)/T_2`

Explanation:

Temperature must be quoted on the `"Kelvin scale"`, and so...

`P_2=(P_1V_1)/T_1xxT_2/V_2`, which will certainly give us an answer with the required units of pressure. Why so?

And thus...........

`P_2=(100*kPaxx73.0*mL)/(273.15*K)xx(353.15*K)/(4530*mL)`

`~=2*kPa`

The pressure change is marked because we increase the volume GREATLY........i.e. more than 50 times the original volume, whereas the temperature is increased only marginally.