5. What is the pressure of a 5.0 L container which contains 3.0 moles of a gas at a temperature of 0°C? Note: Use Ideal gas law. R=8.31 (kPax L/mol x K)

P = 1361.17 KPa

Explanation:

Given data:

Mass of sample = 3.0 mol

Pressure of sample = ?

Volume of gas = 5.0 L

Temperature of gas = 0°C

Solution:

Pressure of gas:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 8.31 KPa.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

0+273 = 273 K

by putting values,

P × 5.0 L = 3.0 mol × 8.31 KPa.L/ mol.K  × 273 K

P = 6805.89 KPa.L/ 5.0 L

P = 1361.17 KPa