A 10.0 l gas cylinder contains nitrogen gas at a pressure of 98.2 PSI at 24°C what will the pressure of the gas be if the cylinder is cooled to 16°C

Answer 1

To solve this problem, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles
R = universal gas constant (0.0821 L.atm/mol.K)
T = temperature in Kelvin

First, we need to convert the initial temperature (24°C) and the new temperature (16°C) to Kelvin:

Initial temperature: 24°C + 273 = 297 K
New temperature: 16°C + 273 = 289 K

Now, we can rearrange the ideal gas law equation to solve for the final pressure:

P1V1/T1 = P2V2/T2

P2 = P1 * (V1/V2) * (T2/T1)

Given:
Initial pressure (P1) = 98.2 PSI
Initial volume (V1) = 10.0 L
Initial temperature (T1) = 297 K
New temperature (T2) = 289 K

The volume of the gas does not change, so V1 = V2

P2 = 98.2 * (10.0/10.0) * (289/297)
P2 = 98.2 * 0.9731 * 0.9714
P2 = 91.0 PSI

Therefore, the pressure of the gas will be approximately 91.0 PSI when the cylinder is cooled to 16°C.