#"2.50 L"# of a gas at #"303 K"# and #"100 kPa"# are heated. The new volume is #"3.75 L"# and the new pressure is #"90.0 kPa"#. What is the new temperature?

1 Answer
Aug 9, 2017

Answer:

#"409 K"# to three significant figures.

Explanation:

This question is an example of the combined gas law. The formula is:

#(P_1V_1)/T_1=(P_2V_2)/T_2#

Organize the data:

Known

#P_1="100 kPa"#

#V_1="2.50 L"#

#T_1="303 K"#

#P_2="90.0 kPa"#

#V_2="3.75 L"#

Unknown

#T_2#

Solution

Rearrange the formula to isolate #T_2#. Insert your data and solve.

#T_2=(P_2V_2T_1)/(P_1V_1)#

#T_2=((90.0color(red)cancel(color(black)("kPa"))xx3.75color(red)cancel(color(black)("L"))xx303"K"))/((100color(red)cancel(color(black)("kPa"))xx2.50color(red)cancel(color(black)("L"))))="409K"# rounded to three significant figures

(#"100 kPa"# is an exact number so it has an infinite number of significant figures.)