If 17 L of a gas at a temperature of 67°C and a pressure of 88.89 atm has its temperature increased to 94°C and its volume decreased to 12 L, what is the new pressure?

Feb 12, 2017

The new pressure will be $\text{140 atm}$.

Explanation:

This is an example of the combined gas law. The equation to use is:

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

where $P$ is the pressure, $V$ is the volume, and $T$ is the temperature in Kelvins. For this question, the temperature in degrees Celsius will be converted to Kelvins by adding $273.15$.

Write what you know.

${P}_{1} = \text{88.89 atm}$
${V}_{1} = \text{17 L}$
${T}_{1} = \text{67"^@"C" + 273.15="340 K}$
${V}_{2} = \text{12 L}$
${T}_{2} = \text{94"^@"C" + 273.15="367 K}$

Write what you don't know: ${P}_{2}$

Solution
Rearrange the equation to isolate ${P}_{2}$, substitute the known values into the equation, and solve.

"P_2=(P_1V_1T_2)/(V_2T_1)

P_2=(88.89"atm"xx17"L"xx367"K")/(12"L"xx340"K")="140 atm" rounded to two significant figures