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?

1 Answer
May 23, 2017

Answer:

#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#