A 75.0-liter canister contains 15.82 moles of argon at a pressure of 546.8 kilopascals. What is the temperature of the canister?

312 K

Explanation:

1) dа ta:

a) V = 75.0 liter

b) n = 15.82 mol

c) p = 546.8 kPa

d) T = ?

2) Formula:

Ideal gas equation: pV = nRT

Where:

n = number of moles V = volume p = absolute pressure T = absolute temperature R = Universal Gas constat: 8.314 kPa - liter / K-mol

3) Solution:

a) Solve the equation for T:

T = pV / (nR)

b) Substitute and compute:

T = 546.8 kPa * 75.0 l iter / (15.82 mol * 8.314 kPa-liter/K-mol) = 312 K

(since the volume is expressed with 3 significant figures, the answer must show also 3 significant figures)