Question

type the correct answer in the box. express your answer to three significant figures. 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? the temperature of the canister is k

Explanation:

Step1: Recall ideal - gas law

$PV = nRT$

Step2: Rearrange for temperature

$T=\frac{PV}{nR}$

Step3: Identify given values

$P = 546.8\ kPa=546800\ Pa$, $V = 75.0\ L = 0.075\ m^{3}$, $n = 15.82\ mol$, $R=8.314\ J/(mol\cdot K)$

Step4: Substitute values

$T=\frac{546800\ Pa\times0.075\ m^{3}}{15.82\ mol\times8.314\ J/(mol\cdot K)}$

Step5: Calculate

$T=\frac{41010\ J}{131.53748\ J/K}\approx311\ K$

Answer:

311 K