Question #cabd7

1 Answer
Jan 6, 2016

They should give the same answer of #V = 3.41 L#, assuming that the amount of gas (i.e., moles) is kept constant.

Explanation:

Using the Ideal Gas Law, #PV = nRT#.

We can take the first scenario, with

#P = 1.03 atm#
#V = 2.20 L#
#R = 0.082057338 (atm L)/(mol K)#
#T = 320.15 K#.

Which yields the following value for #n#:

#n = (PV)/(RT) = (1.03 * 2.20)/(0.082057338 * 320.15)#

#n = 0.0862559 mol#

Using that value for #n# on the latter situation, with

#P = 0.789 atm#
#T = 380.15K#.

We get the following solution for #V#:

#V = (nRT)/P = (0.0862559 * 0.082057338 * 380.15)/0.789#

#V = 3.4102357 L#

Using the 'combined gas law': #(P_1 V_1 )/ T_1 = (P_2 V_2) / T_2#.

We can just plug in the values straight away!

#V_2 = (P_1 V_1)/(T_1) * (T_2)/(P_2)#

#V_2 = (1.03 * 2.20)/(320.15) * (380.15)/(0.789)#

Which, allegedly equals,

#V_2 = 3.4102357 L#, which is as we expected.

I would say the latter method is much faster, however, it's not as 'general' as the Ideal Gas Law.