Question

What pressure in kPa will 2g of argon exert when placed in a 500mL cylinder at 18oC?

Answer 1

using ideal gas equation:

PV = nRT

P = $\frac{nRT}{V}$

where n = no. of moles = ($\frac{2}{40}$ ) = 0.05moles

Volume = 0.5 L

Temperature = 18^oC = 291K

P = $\frac{0.05\times 0.0821 \times 291}{0.5}$

P = 2.38 atm

1 atm = 101.325 kPa

2.38 atm = 241.1535 kPa

Pressure = 241.1535 kPa