Question

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?

Answer 1

`165^oC`

Explanation:

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

`(P_1V_1)/(T_1) = (P_2V_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 (`"L", "K", "atm",`and `"mol"`). The only measurement that needs conversion here is temperature, from `"^oC` to `"K"`. (You will always convert temperature to Kelvin (absolute temperature) when using gas equations).

The Kelvin temperature is

`"K" = 20^(o)C + 273 = 293"K"`

Let's rearrange the combined gas law to solve for the final temperature, `T_2`:

`T_2 = (P_2V_2T_1)/(P_1V_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 `"K"` to `"^oC`:

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