Answer: The pressure exerted by the [tex]CO_2[/tex] gas, in atm is 1.092
Explanation:
According to the ideal gas equation:'
[tex]PV=nRT[/tex]
P = Pressure of the gas = ?
V= Volume of the gas
T= Temperature of the gas = [tex]22.165^0C=(273+22.165)K=295.165K[/tex] (0°C = 273 K)
n= moles of gas = [tex]\frac{\text {given mass}}{\text {Molar mass}}[/tex]
R= Value of gas constant = 0.0821 Latm/K mol
[tex]P=\frac{mRT}{MV}[/tex] as [tex]Density=\frac{mass}{Volume}[/tex]
[tex]P=\frac{dRT}{M}[/tex] where d is density
[tex]P=\frac{1.983g/L\times 0.0821Latm/Kmol\times 295.165K}{44g/mol}=1.092atm[/tex]
Thus pressure exerted by the [tex]CO_2[/tex] gas, in atm is 1.092