Question

if 5.00 l of argon gas is at 0.460 atm and -123°c, what is the volume at stp?
4.19 l
4.94 l
5.06 l
5.49 l
5.97 l

Explanation:

Step1: Convert temperature to Kelvin

Initial temperature \(T_1=- 123^{\circ}C=( - 123 + 273)K = 150K\). Standard temperature \(T_2 = 273K\), standard pressure \(P_2=1atm\). Initial volume \(V_1 = 5.00L\), initial pressure \(P_1=0.460atm\)

Step2: Use the combined gas law \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\)

Rearrange for \(V_2\): \(V_2=\frac{P_1V_1T_2}{P_2T_1}\)
Substitute values: \(V_2=\frac{0.460\times5.00\times273}{1\times150}\)

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Answer:

4.19 L