help me pleaseeeeeee

Question

A typical weather balloon contains approximately 563 L of hydrogen gas when it is released.  If the temperature and pressure at the surface are 288 K and 1.00 atm when the balloon is released, what will its volume be at 32.0 km above the surface, where temperature and pressure are 229 K and 8.57 × 10^-3 atm?

Answer 1

In this problem you have two states; an initial state and a final state

V1 = initial volume = 563 L

P1 = initial pressure = 1.00 atm

T1 = initial temperature = 288K

V2 = final volume = ?

P2 = final pressure = 8.57x10^-3 atm

T2 = final temperature = 229 K

We can now use the Combine Gas Law (not the Ideal Gas Law): P1V1/T1 = P2V2/T2 and solve for V2.

V2 = P1V1T2 / P2T1

V2 = (1.00)(563)(229) / (8.57x10^-3)(288)

V2 = 5.22x10⁴ L (3 sig. figs.)

Answer 2

Use the gas law pV = nRT

V = 563 T = 288 p = 1, R constant, giving n = 1.95 / R

When p = 8.57 x 10^-3 and T = 229, V = nRT/P = 1,95 x 229 / (8.57 x 10^ -3) = 52106 L