Answer:
The additional information required to solve this problem is the initial volume.
the final pressure P₂ of the gas is 1.108 atm
Explanation:
Given that :
A sample of gas at initial temperature [tex]T_1 = 12.0^0 \ C[/tex] = (12+273)K = 285 K
Pressure (P₁) = 1.06 atm
Initial Volume (V₁) = unknown ???
Final Volume (V₂) = 2.30 L
final temperature [tex]T_2 = 24.9^0 \ C[/tex] = (24.9 +273)K = 297.9 K
Find the final Pressure (P₂)
The relation between: Pressure, Volume and Temperature can be gotten from the ideal gas equation :
PV = nRT
The Ideal Gas Equation is also reduced to the General Gas Law or the combined Gas Law by assuming that n= 1 .
From ; PV = nRT
[tex]\frac{PV}{T} = R \ \ ( constant) \ if \ n=1[/tex]
∴ [tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = \frac{P_3V_3}{T_3}...= \frac{P_nV_n}{T_n} \ \ \ ( n \ constant)[/tex]
The additional information required to solve this problem is the initial volume.
This expression is a combination of Boyle's Law and Charles Law. From the combined Gas Law , it can be deduced that at constant volume, the pressure of a given mass(mole) of gas varies directly with absolute temperature.
∴ [tex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/tex] if n & Volume (V) are constant .
[tex]P_2 = \frac{1.06*297.9}{285}[/tex]
P₂ = 1.108 atm
Thus, the final pressure P₂ of the gas is 1.108 atm
