Answer:
[tex]\rho =1.96\frac{g}{L}[/tex]
Explanation:
Hello there!
In this case, since this imaginary gas can be modelled as an ideal gas, we can write:
[tex]PV=nRT[/tex]
Which can be written in terms of density and molar mass as shown below:
[tex]\frac{P}{RT} =\frac{n}{V} \\\\\frac{P}{RT} =\frac{m}{MM*V}\\\\\frac{P*MM}{RT} =\frac{m}{V}=\rho[/tex]
Thus, by computing the pressure in atmospheres, the resulting density would be:
[tex]\rho = \frac{165/760 atm * 314.2 g/mol}{0.08206\frac{atm*L}{mol*K}*425K} \\\\\rho =1.96\frac{g}{L}[/tex]
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