Answer :
Answer : The final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa
Explanation :
First we have to calculate the initial pressure of gas.
As we know that, R-134a is 1,1,1,2-Tetrafluoroethane.
Using ideal gas equation:
[tex]PV=nRT\\\\PV=\frac{w}{M}RT[/tex]
where,
P = pressure of gas = ?
V = volume of gas = [tex]1.115m^3[/tex]
T = temperature of gas = [tex]-30^oC=273+(-30)=243K[/tex]
R = gas constant = [tex]8.314m^3Pa/mole.K[/tex]
w = mass of gas = 10 kg = 10000 g
M = molar mass of R-134a gas = 102.03 g/mole
Now put all the given values in the ideal gas equation, we get:
[tex]P\times 1.115m^3=\frac{10000g}{102.03g/mole}\times (8.314m^3Pa/mole.K)\times (243K)[/tex]
[tex]P=177587.9687Pa=177.6kPa[/tex]
Thus, the initial pressure of gas is, 177.6 kPa
Now we have to calculate the final temperature of gas.
Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.
[tex]P\propto T[/tex]
or,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 177.6 kPa
[tex]P_2[/tex] = final pressure of gas = 200 kPa
[tex]T_1[/tex] = initial temperature of gas = [tex]-30^oC=273+(-30)=243K[/tex]
[tex]T_2[/tex] = final temperature of gas = ?
Now put all the given values in the above equation, we get:
[tex]\frac{177.6kPa}{243K}=\frac{200kPa}{T_2}[/tex]
[tex]T_2=273.6K[/tex]
Thus, the final temperature an the initial pressure of gas is, 273.6 K and 177.6 kPa