# A sample of a gas has a pressure of "0.370 atm" at a temperature of "5.00"^@"C". What will the pressure be if the temperature is lowered to "0"^@"C"?

May 12, 2017

The pressure at $\text{273.15 K}$ $\left({0}^{\circ} \text{C}\right)$ will be $\text{0.363 atm}$.

#### Explanation:

Gay-Lussac's gas law will be used to solve this problem. It states that the pressure of a gas is directly proportional to its Kelvin temperature, as long as mass and volume are held constant. This means that if the volume increases, so does the temperature and vice-versa. The equation for this law is:

${P}_{1} / {T}_{1} = {P}_{2} / {T}_{2}$

The given data has temperature in degrees Celsius, so it needs to be converted to Kelvins. This is done by adding $\text{273.15}$ to the Celsius temperature.

Given

${P}_{1} = \text{0.370 atm}$

${T}_{1} = \text{5.00"^@"C" + 273.15="278.15 K}$

${T}_{2} = \text{0.00"^@"C + 273.15 ="273.15 K}$

Unknown:

${P}_{2}$

Solution
Rearrange the equation to isolate ${P}_{2}$. Insert the given data and solve.

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

P_2=(0.370color(white)(.)"atm"xx273.15color(red)cancel(color(black)("K")))/(278.15color(red)cancel(color(black)("K")))="0.363 atm" (rounded to three significant figures)