Ideal Gas Law & Combined Gas Law Problem

[Ginerva123]

[SOLVED] Gas laws

Homework Statement

A balloon containing 2000m^3 of helium gas at 1.00atm and a temperature of 15.0 C rises from ground level to an altitude at which the atmospheric pressure is only 0.900atm. Assume the helium behaves like an ideal gas and the balloon's ascent is too rapid to permit much heat exchange with the surrounding air. Calculate both the volume and the temperature of the gas at the higher altitude.

Homework Equations

Ideal gas equation; combined gas law.

The Attempt at a Solution

Not really sure how to procede... If I had the new temperature, i could find the new volume, and vice versa, but without either, I'm stuck. Also, if there is little heat exchange with the surrounding air, how does the temperature change?

 

[Hootenanny]

You know that the process is adiabatic, you also know the initial and final pressures.

 

[Moetasim]

The ideal gas law and combined gas law can be used to solve this problem. First, we can use the ideal gas law, PV = nRT, to calculate the number of moles of helium gas present in the balloon at ground level. We know that the pressure is 1.00 atm, the volume is 2000m^3, and the temperature is 15.0 C (288 K). We can rearrange the equation to solve for n, the number of moles:

n = PV/RT
n = (1.00 atm)(2000m^3)/(0.0821 L*atm/mol*K)(288 K)
n = 86.8 moles

Now, we can use the combined gas law, P1V1/T1 = P2V2/T2, to calculate the new volume and temperature of the gas at the higher altitude. We know that the pressure has decreased to 0.900 atm, and we can use the calculated number of moles (86.8 moles) to solve for the new volume. We also know that the temperature remains constant at 15.0 C (288 K).

P1V1/T1 = P2V2/T2
(1.00 atm)(2000m^3)/(288 K) = (0.900 atm)(V2)/(288 K)
V2 = (1.00 atm)(2000m^3)(288 K)/(0.900 atm)(288 K)
V2 = 2222m^3

Therefore, the volume of the gas at the higher altitude is 2222m^3.

To calculate the new temperature, we can use the same equation, but solve for T2:

P1V1/T1 = P2V2/T2
(1.00 atm)(2000m^3)/(288 K) = (0.900 atm)(2222m^3)/T2
T2 = (0.900 atm)(288 K)(2000m^3)/(1.00 atm)(2222m^3)
T2 = 259.2 K

Therefore, the temperature of the gas at the higher altitude is 259.2 K (15.2 C).

In conclusion, using the ideal gas law and combined gas law, we have calculated that at the higher altitude, the volume of the gas increases to 2222m^3 and