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

User Venkat Reddy
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1 Answer

16 votes
16 votes

Answer:

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,
image

So,


image

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

User VSP
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