Answer :
Answer:
[tex]0.16098\times 10^{-3}\ g/cm^3[/tex]
Explanation:
P =Pressure = 9846 Pa
V = Volume
n = Amount of substance = 1
T = Temperature = 21.1°C
[tex]\rho[/tex] = Density
R = Gas constant = 8.314 J/mol K
M = Molar mass of argon = 40 g/mol
From ideal gas law we have the relation
[tex]PV=nRT[/tex]
Multiply density on both sides
[tex]PV\rho=nR\rho T\\\Rightarrow PM=nR\rho T\\\Rightarrow \rho=\dfrac{PM}{nRT}\\\Rightarrow \rho=\dfrac{9846\times 40\times 10^{-3}}{8.314\times (21.1+273.15)}\\\Rightarrow \rho=0.16098\ kg/m^3\\\Rightarrow \rho=0.16098\times 10^{-3}\ g/cm^3[/tex]
The density of argon gas is [tex]0.16098\times 10^{-3}\ g/cm^3[/tex]