# A certain gas occupies a volume of 550.0 mL at STP. What would its volume be at 27°C and 125.0 kPa?

Nov 17, 2015

The volume of the gas at $\text{300 K}$ and $\text{125.0 kPa}$ will be $\text{0.48 L}$

#### Explanation:

Use the combined gas law to solve this problem. The equation is $\frac{{V}_{1} {P}_{1}}{T} _ 1 = \frac{{V}_{2} {P}_{2}}{T} _ 2$

$\text{STP"="273.15 K and 100 kPa}$

Given
${V}_{1} = 550.0 \text{mL"xx(1"L")/(1000"mL")="0.5500 L}$
${P}_{1} = \text{100 kPa}$
${T}_{1} = \text{273.15 K}$
${P}_{2} = \text{125.0 kPa}$
${T}_{2} = {27}^{\text{o""C""+273.15=300"K}}$

Unknown
${V}_{2}$

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

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

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

V_2=((0.5500"L"xx100"kPa"xx300"K"))/((273.15"K"xx125.0"kPa"))="0.48 L" (rounded to two significant figures.