Answer:
[tex] V=\frac{1.82 mol *0.082 \frac{atm L}{mol K} *303 K}{0.595 atm}=76.03 L[/tex]
Step-by-step explanation:
For this case we can use the Ideal gas law given by:
[tex] PV =nRT[/tex]
Where:
P the pressure in atm
V the volume in liters
n the amount of mass in moles
[tex]R=0.082 \frac{atm L}{mol K}[/tex] a constant
T the temperature in Kelvin
And if we solve for the value of V we got:
[tex] V =\frac{nRT}{P}[/tex]
For this case we know that P = 452 Torr, but we need to convert this into atm like this:
[tex] P = 452 Torr *\frac{1 atm}{760 Torr}=0.595 Torr[/tex]
The other variables are known:
n =1.82 mol, T =303 K
And now we have everything to replace:
[tex] V=\frac{1.82 mol *0.082 \frac{atm L}{mol K} *303 K}{0.595 atm}=76.03 L[/tex]
