Answer:
![22.88\text{atm}](https://img.qammunity.org/2023/formulas/chemistry/high-school/8q9ex14j9xoa3tf2eme5cc1zkwzixuap4m.png)
Explanations:
In order to get the pressure of the gas, we will use the ideal gas equation expressed as:
![PV=\text{nRT}](https://img.qammunity.org/2023/formulas/physics/college/ani9pbfsi66dtnimlr7m7wedu5x6xa0heo.png)
• P is the ,pressure ,of the gas (in atm)
• V is the v,olume ,of the gas (in L)
• n is the ,number of moles, of gas (in moles)
• R is the ,gas constant
• T is the ,temperature, in kelvin
Given the following parameters:
![\begin{gathered} n=0.0925\text{moles} \\ R=0.0821L\cdot atm/mole\cdot K \\ V=152mL=0.152L \\ T=185^0C=185+273=458K \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/high-school/f0mwk2rjxwoxw2k7jyac2vr92kuzhi7gkl.png)
Substitute the given parameters into the formula:
![\begin{gathered} P=\frac{\text{nRT}}{V} \\ P=\frac{0.0925\cancel{\text{moles}}*0.0821\frac{\cancel{L}\cdot atm}{\cancel{\text{mole}\cdot K}}*458\cancel{K}}{0.152\cancel{L}} \\ P=(0.0925*0.0821atm*458)/(0.152) \\ P=(3.4781665)/(0.152)\text{atm} \\ P=22.88\text{atm} \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/high-school/w6nfy52rc2i9uvivx7fas5tjnjxu0b72hl.png)
Hence the pressure of the gas is 22.88atm