Answer:
V = 177.4 L.
Explanation:
Hello there!
In this case, since this gas can be assumed as ideal due to the given data, we can use the following equation:
[tex]PV=nRT\\\\PV=\frac{m}{MM}RT[/tex]
Thus, by solving for volume we obtain:
[tex]V=\frac{mRT}{MM*P}[/tex]
So we can plug in the temperature in Kelvins (537 K), the pressure in atmospheres (0.404 atm) and the molar mass (54 g/mol) to obtain:
[tex]V=\frac{87.8g*0.08206\frac{atm*L}{mol*K}*537K}{54g/mol*0.404atm}\\\\V=177.4L[/tex]
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