# 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?

Aug 23, 2017

Well for a given molar quantity of gas, $\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$

#### Explanation:

Temperature must be quoted on the $\text{Kelvin scale}$, and so...

${P}_{2} = \frac{{P}_{1} {V}_{1}}{T} _ 1 \times {T}_{2} / {V}_{2}$, which will certainly give us an answer with the required units of pressure. Why so?

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

${P}_{2} = \frac{100 \cdot k P a \times 73.0 \cdot m L}{273.15 \cdot K} \times \frac{353.15 \cdot K}{4530 \cdot m L}$

$\cong 2 \cdot k P a$

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.