Question

What mass of argon occupies 4.3 l at 70 kpa and 20 c?

Answer 1

Let us assume that Argon gas is acting Ideally,' Then, According to Ideal Gas Equation,

P V = n R T ----- (1)
Also,
moles = n = Mass / M.mass = m / M

Substituting value of n in equation 1,

P V = (m/M) R T

Solving for mass 'm',

m = P V M / R T ----- (2)
Data Given;
P = 70 kPa = 0.69 atm

V = 4.3 L

T = 20 °C + 273 = 293 K

R = 0.0821 atm.L.mol⁻¹.K⁻¹

M = 39.94 g.mol⁻¹

Now, putting values in equation 2,

m = (0.69 atm × 4.3 L × 39.94 g.mol⁻¹) ÷ (0.0821 atm.L.mol⁻¹.K⁻¹ × 293 K)

m = 4.89 g of Argon