A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to

Question

3 years ago

13

A rigid, 2.50 L bottle contains 0.458 mol He. The pressure of the gas inside the bottle is 1.83 atm. If 0.713 mol Ar is added to

the bottle and the pressure increases to 2.05 atm, what is the change in temperature of the gas mixture? The initial temperature of the gas is 122 K. The final temperature of the gas is K.

Chemistry

2 answers:

liraira [26] · Answer 1 · 2022-02-11T11:50:19+03:00

liraira [26]3 years ago

8 0

Answer:

Initial is 122. The final is 53.3

Alex17521 [72] · Answer 2 · 2022-02-11T09:30:58+03:00

137 K

The volume is constant, so you can use <em>Gay-Lussac’s Pressure-Temperature Law </em>to calculate the new temperature (you don’t have to use the number of moles).

P1/T1 = P2/T2

Solve for T2: T2= T1 x P2/P1

P1 = 1.83 atm; T1 = 122 K

P2 = 2.05 atm; T2 = ?

∴ T2 = 122 K x (2.05 atm)/(1.83 atm) = 137 K

This result makes sense. Temperature is directly proportional to pressure. You increased the pressure by about 10 %, so the temperature increased by about 10 %.