Assuming the gas behaves as an ideal gas, we can use the Ideal Gas Law to calculate the pressure:
![\begin{gathered} PV=nRT \\ P=(nRT)/(V) \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/9i4xak22ovy72n5lpy1jg0gogcxed4iqgo.png)
We need to use T in absolute terms, so we need to convert it to Kelvin:
![T=(28+273.15)\; K=301.15\; K](https://img.qammunity.org/2023/formulas/chemistry/college/jc2algtfhoq0e9jiusxmyl3sxcp64hvb6h.png)
Since we have the volume in L and we want the pressure in atm, we can use the following unit for the R constant:
![R\approx0.082057\; atm\cdot L\cdot K^(-1)\cdot mol^(-1)](https://img.qammunity.org/2023/formulas/chemistry/college/vy97okvmkig0yg44iln1r1p2hh0myc4476.png)
Using the these and the other given values:
![\begin{gathered} V=7.32\; L \\ n=0.448\; mol \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/6shltl6ixw02z7fmcr71f0xa01k9qow8uz.png)
We have:
![\begin{gathered} P=(nRT)/(V) \\ P=(0.448mol\cdot0.082057atm\cdot L\cdot K^(-1)\cdot mol^(-1)\cdot301.15K)/(7.32L) \\ P=(0.448\cdot0.082057\cdot301.15)/(7.32)\; atm \\ P=1.51239\ldots\; atm \\ P=1.51\; atm \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/nx7bxpz4177yu88vc7xwfumojolzz6ildh.png)
So, the pressure is approcimately 1.51 atm.