Respuesta :
Answer : The total change of volume is, 41.883 liters.
Explanation :
R134a is a 1,1,1,2-tetrafluoroethane. It is a hydro-fluorocarbon and haloalkane gaseous refrigerant.
First we have to calculate the volume at [tex]-20^oC[/tex].
Using ideal gas equation:
[tex]PV=nRT\\\\PV=\frac{w}{M}\times RT[/tex]
where,
n = number of moles
w = mass of R134a = 257 g
P = pressure of the gas = 60 Kpa
T = temperature of the gas = [tex]-20^oC=273+(-20)=253K[/tex]
M = molar mass of R134a = 102.03 g/mole
R = gas constant = 8.314 Kpa.L/mole.K
V = initial volume of gas
Now put all the given values in the above equation, we get :
[tex](60Kpa)\times V=\frac{257g}{102.03g/mole}\times (8.314Kpa.L/mole.K)\times (253K)[/tex]
[tex]V=88.305L[/tex]
Now we have to calculate the volume at [tex]100^oC[/tex] by using Charles's law.
Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
[tex]V\propto T[/tex]
or,
[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of gas = 88.305 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = 253 K
[tex]T_2[/tex] = final temperature of gas = [tex]100^oC=273+100=373K[/tex]
Now put all the given values in the above formula, we get the final volume of the gas.
[tex]\frac{88.305L}{V_2}=\frac{253K}{373K}[/tex]
[tex]V_2=130.188L[/tex]
Now we have to calculate the total change of volume.
[tex]V_2-V_1=130.188-88.305=41.883L[/tex]
Therefore, the total change of volume is, 41.883 liters.
