The gas in a closed container has a pressure of #3.00 * 10^2# #kPa# at #30°C# #(303 K)#. What will the pressure be if the temperature is lowered to #-172°C# #(146 K)#?

1 Answer
Meave60
Feb 22, 2016

The final pressure will be 145 kPa.

Explanation:

This is an example of Gay-Lussac's law, which states that the pressure of a given gas is directly proportional to its Kelvin temperature, when held at a constant volume. The equation that represents this law is #P_1/T_1=P_2/T_2#.

Given/Known
#P_1=3.00xx10^2"kPa"#
#T_1="303 K"#
#T_2=-127^@"C+273.15=146 K"# **

Unknown
#P_2#

Solution
Rearrange the equation to isolate #P_2#. Substitute the given/known values into the equation and solve.

#P_1/T_1=P_2/T_2#

#P_2=(P_1T_2)/T_1#

#P_2=(3.00xx10^2"kPa" xx 146cancel"K")/(303cancel"K")="145 kPa"#

**Kelvin temperature changed to 146 K, which is #(-127^@ "C"+273.15)#

Excellent Resource for the gas laws: http://chemistry.bd.psu.edu/jircitano/gases.html

Related questions
See all questions in Gay Lussac's Law
Impact of this question
15271 views around the world
You can reuse this answer
Creative Commons License