Answer : The pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.
Solution :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
![\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
where,
= initial pressure of gas = 3 atm
= final pressure of gas = ?
= initial volume of gas = 1.40 L
= final volume of gas = 0.950 L
= initial temperature of gas = ![35^oC=273+35=308K](https://tex.z-dn.net/?f=35%5EoC%3D273%2B35%3D308K)
= final temperature of gas = ![0^oC=273+0=273K](https://tex.z-dn.net/?f=0%5EoC%3D273%2B0%3D273K)
Now put all the given values in the above equation, we get the final pressure of gas.
![\frac{3atm\times 1.40L}{308K}=\frac{P_2\times 0.950L}{273K}](https://tex.z-dn.net/?f=%5Cfrac%7B3atm%5Ctimes%201.40L%7D%7B308K%7D%3D%5Cfrac%7BP_2%5Ctimes%200.950L%7D%7B273K%7D)
![P_2=3.918atm](https://tex.z-dn.net/?f=P_2%3D3.918atm)
Therefore, the pressure of gas will be, 3.918 atm and the combined gas law is used for this problem.