# A 3.50-L gas sample at 20°C and a pressure of 86.7 kPa expands to a volume of 8.00 L. The final pressure of the gas is 56.7 kPa. What is the final temperature of the gas, in degrees Celsius?

May 23, 2017

${165}^{o} C$

#### Explanation:

We can use here the combined gas law, relating the temperature, pressure, and volume of a gas with a constant quantity:

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

If you're using the ideal-gas equation, which we're NOT using here, you would have to convert each measurement into the appropriate units ($\text{L", "K", "atm} ,$and $\text{mol}$). The only measurement that needs conversion here is temperature, from "^oC to $\text{K}$. (You will always convert temperature to Kelvin (absolute temperature) when using gas equations).

The Kelvin temperature is

$\text{K" = 20^(o)C + 273 = 293"K}$

Let's rearrange the combined gas law to solve for the final temperature, ${T}_{2}$:

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

Plugging in known values, we can find the final temperature:

T_2 = ((56.7"kPa")(8.00"L")(293"K"))/((86.7"kPa")(3.50"L")) = 438"K"

Lastly, we'll convert back from $\text{K}$ to "^oC:

438"K" - 273 = color(blue)(165^oC