The pressure of the gas is about 11.1 atm
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Further explanation</h3>
The Ideal Gas Law and Work formula that needs to be recalled is:
![\boxed {PV = nRT}](https://tex.z-dn.net/?f=%5Cboxed%20%7BPV%20%3D%20nRT%7D)
![\boxed { W = P \Delta V }](https://tex.z-dn.net/?f=%5Cboxed%20%7B%20W%20%3D%20P%20%5CDelta%20V%20%7D)
<em>where:</em>
<em>W = Work ( J )</em>
<em>P = Pressure ( Pa )</em>
<em>V = Volume ( m³ )</em>
<em>n = number of moles ( moles )</em>
<em>R = Gas Constant ( 8.314 J/mol K )</em>
<em>T = Absolute Temperature ( K )</em>
Let us now tackle the problem !
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<u>Given:</u>
amount of substance = n = 1.50 mol
volume of gas = V = 3.30 L
temperature of gas = T = 25 + 273 = 298 K
gas constant = R = 0.0821 L.atm/mol.K
<u>Asked:</u>
pressure of gas = P = ?
<u>Solution:</u>
![PV = n R T](https://tex.z-dn.net/?f=PV%20%3D%20n%20R%20T)
![P = n R T \div V](https://tex.z-dn.net/?f=P%20%3D%20n%20R%20T%20%5Cdiv%20V)
![P = 1.50 \times 0.0821 \times 298 \div 3.30](https://tex.z-dn.net/?f=P%20%3D%201.50%20%5Ctimes%200.0821%20%5Ctimes%20298%20%5Cdiv%203.30)
![P \approx 11.1 \texttt{ atm}](https://tex.z-dn.net/?f=P%20%5Capprox%2011.1%20%5Ctexttt%7B%20atm%7D)
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Conclusion :</h3>
The pressure of the gas is about 11.1 atm
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Learn more</h3>
![\texttt{ }](https://tex.z-dn.net/?f=%5Ctexttt%7B%20%7D)
<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Pressure