Answer: At a pressure of 15.613 atm a 5.32 mol [tex]Cl_{2}[/tex] sample occupy 5.08 L at 181.59 K.
Explanation:
Given: Moles = 5.32 mol
Volume = 5.08 L
Temperature = 181.59 K
Formula used to calculate the pressure is as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.
[tex]PV = nRT\\P \times 5.08 L = 5.32 mol \times 0.0821 L atm/mol K \times 181.59 K\\P = \frac{5.32 mol \times 0.0821 L atm/mol K \times 181.59 K}{5.08 L}\\= 15.613 atm[/tex]
Thus, we can conclude that at a pressure of 15.613 atm a 5.32 mol [tex]Cl_{2}[/tex] sample occupy 5.08 L at 181.59 K.
