Question

What is the preasure in atmospheres of 20 mol of nitrogen gas in 36.2 L cylinder at 25 degrees C?

Answer 1

P = 13.5 atm

Step-by-step explanation:

Given that

No. of moles, n = 20 moles

Volume of nitrogen gas = 36.2 L

Temperature = 25°C = 298 K

We need to find the pressure of the gas. Using the ideal gas equation

PV = nRT

Where

R is gas constant,
`R=0.082057\ L-atm/K-mol`

So,

`P=(nRT)/(V)\\\\P=(20* 0.082057* 298)/(36.2 )\\\\P=13.5\ atm`

so, the pressure of the gas is equal to 13.5 atm.