# A 1.15 mol sample of carbon monoxide gas has a temperature of 27°C and a pressure of 0.300 atm. If the temperature were lowered to 17°C, at constant volume, what would be the new pressure?

Oct 10, 2016

The new pressure will be 0.29 atm .

#### Explanation:

This is an example of Gay-Lussac's gas law, which states that the pressure of a given amount of gas kept at constant volume is directly proportional to the temperature in Kelvins.

The equation for this gas law is $\frac{{P}_{1}}{{T}_{1}} = \frac{{P}_{2}}{{T}_{2}}$. Notice that the number of moles is not a variable in the equation. The number of moles was given to indicate that the $\text{1.15 mol CO"_2}$ does not change, and is the "given amount of gas."

Determine the known and unknown values.

Known
${P}_{1} = \text{0.300 atm}$
${T}_{1} = \text{27"^@"C"+"273.15"="300 K}$
${T}_{2} = \text{17"^@"C"+"273.15"="290 K}$

Unknown
${P}_{2}$

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

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

P_2=((0.300"atm")xx(290cancel"K"))/(300cancel"K")="0.29 atm"