Question

If 0.450 mol of helium gas occupies 5.50 L at 25 C, what is the pressure in the atmosphere?

Answer 1

Let us consider the helium gas behaves like ideal gas. Thus it obeys the Boyle's law.

According to the Boyle's law, at temperature T kelvin if the pressure and volume of the gas are P and V respectively then,

PV = nRT where n = number of moles of gas molecules and R = universal gas constant.

n = 0.450 moles

R = 0.082 L atm K^—1mol^—1.

V = 5.50 L

T = 25 degree celsius = (273+25) Kelvin = 298 K

Thus from the Boyle's law

The preessure P = nRT/ V = ( 0.450 mole) *(0.082 L atm K^—1 mole^—1)*( 298 K) / 5.50 L = 1.99 atm.

The pressure of the gas is 1.99 atm.