A rubber balloon containing 1.0 L of a gas is carried from the top of a mountain to the bottom of the mountain

A rubber balloon containing 1.0 L of a gas is carried from the top of a mountain to the bottom of the mountain, where its volume is measured as 0.85 L at standard pressure. Assuming that there was no temperature change during the trip, what was the pressure at the top of the mountain?

Answer:
PV=nRT, so if you divide case 1 (top of mountain) with case 2 (bottom) you have:
P1*V1/(P2*V2) = n1*R*T1/(n2*R*T2)
since R is a constant, you didn't leak any gas from the balloon, and temperature didn't change, then you get:
P1*V1/(P2*V2) = 1
or
P1*V1=P2*V2
solving for P1, the initial pressure up on the mountain:
P1=P2*V2/V1
P1=1atm*0.85L/1.0L = 0.85atm (assuming a 1 atm standard pressure)
Ideal gas law says pv=nrt. R is always constant, n is constant for this problem because of the balloon, and you say t is constant. Therefore, P1*V1 = NRT(constant) = P2*T2. So, plug in 1.0 L for V1, 0.85 L for V2, and 1 atmosphere for P2 (assuming you want the answer in atmospheres) and solve for P1!

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