# The volume of a sample of a gas at 273°C is 200.0 L. If the volume is decreased to 100.0 L at constant pressure, what will be the new temperature of the gas?

Nov 3, 2015

${T}_{2} = \text{273 K}$

#### Explanation:

Charles' Law states that when pressure is held constant, the temperature and volume of a gas are directly proportional, so that if one goes up, so does the other. Note: The temperature needs to be in Kelvins.

The equation for Charles' Law is ${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$

${V}_{1} = \text{200.0 L}$
${T}_{1} = \text{273"^"o""C+273=546 K}$
${V}_{2} = \text{100.0 L}$
${T}_{2} = \text{?}$

Solution
Rearrange the equation to isolate ${T}_{2}$ and solve.

${T}_{2} = \frac{{V}_{2} {T}_{1}}{V} _ 1$

${T}_{2} = \left(100.0 \cancel{\text{L"xx546"K")/(200cancel"L}}\right) =$

${T}_{2} = \text{273 K}$