If #3 L# of a gas at room temperature exerts a pressure of #7 kPa# on its container, what pressure will the gas exert if it the container's volume changes to #2/3 L#?

1 Answer
May 15, 2016

Answer:

#P_2= (P_1V_1)/V_2 = (7kPa * 3L)/(2/3 L)=31.5kPa#

Explanation:

This question is asking us to use things we know about gasses to find the final condition after a change of one of the variables. The ideal gas law can help us:

#PV=nRT#

Where #P# is pressure the gas exerts on its container, #V# is the volume of the container, #n# is the number of moles of gas particles, #R# is the universal gas constant, and #T# is the temperature. In our question, the only things that are changing are the pressure and the volume which means that everything else is constant, i.e.

#PV = "constant"#

Since we have two conditions, call them 1 and 2, we can use this expression to solve for them, i.e. in our initial conditions we have

#P_1V_1 = "constant"#

and in our final state we have

#P_2V_2 = "constant"#

the trick is to understand that the constant is the same in both cases (none of #n#, #R#, or #T# are changing), which gives us:

#P_1V_1 = P_2V_2#

We can solve this for #P_2# since that is what the question is asking us for:

#P_2= (P_1V_1)/V_2 = (7kPa * 3L)/(2/3 L)=31.5kPa#