A75.0 liter canister contains 15.82 moles of argon at a pressure of 546.8 kilopascals. what is the temperature of the canister?

Pressure of argon = 546.8 kPa

Conversion factor: 1 atm = 101.325 kPa

Pressure of argon = 546.8 kPa x 1 atm/101.325 kPa = 5.4 atm

Moles of argon = 15.82

Volume of argon = 75.0 L

According to Ideal gas law,

PV = nRT

where P is the pressure, V is the volume , n is the number of moles, R is the universal gas constant, and T is the temperature

T = PV/nR = (5.4 atm x 75.0 L) / (15.82 x 0.0821 L.atm.mol⁻¹K⁻¹)

T = 311.82 K

Hence the temperature of the canister is 311.82 K.

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