Question

A balloon with a volume of 0.851 L holds 0.783 mol He gas at a temperature of 150.0 K. The temperature then increases to 305 K, and the pressure is changed to 0.800 atm. The original pressure of the gas was ______ atm. The final volume of the balloon is _____ L.

Answer 1

11.3atm       24.5L that's is the correct answer 

Answer 2

Answer:

Original pressure = P1 = 11.3 atm

Final volume = V2 = 24.5 L

Explanation:

Given data:

Initial conditions:

Volume V1 = 0.851 L

Temperature T1 = 150.0 K

Moles of He n1 = 0.783

From ideal gas equation:

$PV = nRT\\\\P = \frac{nRT}{V}$

Substituting the initial V, T and n using R = 0.0821 L.atm/mol-K we get:

$P1 = \frac{0.783*0.0821*150.0}{0.851} = 11.33 atm$

Final conditions:

T2 = 305 K

P2 = 0.800 atm

Since the number of moles of He will remain constant we can write:

$\frac{P1V1}{T1} = \frac{P2V2}{T2} \\\\V2 = \frac{P1*V1*T2}{T1*P2} = \frac{11.33*0.851*305}{150*0.800} =24.50 L$