8.00 L of a gas is collected at 60.0°C. What will be its volume upon cooling to 30.0°C?

1 Answer

The gas' new volume will be #V_2 = 7.28# L.

This problem is a simple application of Charles' law

#V_1/T_1 = V_2/T_2#,

which states that the volume of a gas is directly proportional to its temperature(measured in #K# - very important).

So, we have an initial volume equal to #V_1 = 8.00L#, an initial temperature of #T_1 = (273.15+60) = 333.15K#, and a final temperature of #T_2 = (273.15 + 30) = 303.15K#, which gets us

#V_2 = T_2/T_1 * V_1 = 303.15/333.15 * 8.00 = 7.28L# -> a decrease in temperature is correlated with a decrease in volume, just as expected.

Note, however, what would have happened with degrees Celsius instead of K:

#V_2 = 30/60 * 8.00 = 4.00 L#, a result significantly different...