9) A 22.8 mL volume of argon gas is collected at 48.0o C. At constant pressure, what volume would this same sample of gas occupy at standard temperature?

19.39 mL.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have different values of V and T:

(V₁T₂) = (V₂T₁)

Knowing that:

V₁ = 22.8 mL, T₁ = 48°C + 273 = 321 K,

V₂ = ? mL, T₂ = 273 (standard temperature = 273 K),

Applying in the above equation

(V₁T₂) = (V₂T₁)

∴ V₂ = (V₁T₂) / (T₁) = (22.8 mL) (273 K) / (321 K) = 19.39 mL.