Question

A gas occupies a volume of 120. mL at 27°C. At what temperature in °C would the volume be 80.0mL?

Answer 1

To answer this question we will assume the following:

• The amount of gas is constant, that is, the moles do not change.

,

• The pressure remains constant.

,

• The gas behaves like an ideal gas.

With these assumptions, we can apply Charles's law which tells us:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

Where,

V1 is the initial volume of the gas in liters, 120mL=0.120L

T1 is the initial temperature in Kelvin, 27°C+273.15K=300.15K

V2 is the final volume of the gas in liters, 80.0mL=0.080L

T2 is the final temperature in Kelvin, unknown.

We clear T2 and replace the known data:

$T_2=\frac{T_1}{V_1}\times V_2$$\begin{gathered} T_2=\frac{300.15K}{0.120L}\times0.080L=200.1K \\ T_2=-73.1\degree C \end{gathered}$

Answer: The gas would have a volume of 80.0mL at -73.1°C