The temperature of the canister is 311.82(OR 312)K
Given:
volume of canister = 75.0-liter
moles of argon = 15.82 moles
pressure = 546.8 kilopascals
To Find:
temperature of the canister
Solution: Pressure of argon = 546.8 kPa
Conversion factor: 1 atm = 101.325 kPa
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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