Answer:
0.081atm
Explanations:
According to general gas equation expressed as:
![(P_1V_1)/(T_1)=(P_2V_2)/(T_2)](https://img.qammunity.org/2023/formulas/chemistry/college/sbnuxjrz17dtnno6u5lwxy41so88n95l5v.png)
where:
• P1 and P2 are the ,initial and final pressure
,
• V1 and V2 are the ,initial and final volume
,
• T1 and T2 are the i,nitial and final temperature
Given the following parameters
![\begin{gathered} V_1=2.00L \\ V_2=4.50L \\ T_1=-20.0^0C+273=253K \\ T_2=57^0C+273=330K \\ P_1=0.140atm \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/t124qvgv74c4pr3oj54k3x653kr6ngddq0.png)
Required
Final Pressure P2
Substitute the given parameters into the formula
![P_2=(P_1V_1T_2)/(V_2T_1)](https://img.qammunity.org/2023/formulas/chemistry/college/sfbmnouvwldns5lfnkojj2hxhwry1lby16.png)
Substitute the given following parameters into the formula
![\begin{gathered} P_2=(0.14*2.00L*330K)/(4.50L*253K) \\ P_2=(92.4)/(1138.5) \\ P_2=0.081atm \end{gathered}](https://img.qammunity.org/2023/formulas/chemistry/college/d8bd9ihoc3j0els0r43xm9o3ej0zq089mk.png)
Hence the final pressure of the gas sample is 0.081atm