Question

A piston-cylinder device contains 257 grams of R134a at 60 kPa and -20°C. The R134a is heated until its temperature reaches 100°C. Determine the total change of volume.

Answer 1

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
`-20^oC`.

Using ideal gas equation:

`PV=nRT\\\\PV=(w)/(M)* RT`

where,

n = number of moles

w = mass of R134a = 257 g

P = pressure of the gas = 60 Kpa

T = temperature of the gas =
`-20^oC=273+(-20)=253K`

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 :

`(60Kpa)* V=(257g)/(102.03g/mole)* (8.314Kpa.L/mole.K)* (253K)`

`V=88.305L`

Now we have to calculate the volume at
`100^oC` 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.

`V\propto T`

or,

`(V_1)/(V_2)=(T_1)/(T_2)`

where,

`V_1` = initial volume of gas = 88.305 L

`V_2` = final volume of gas = ?

`T_1` = initial temperature of gas = 253 K

`T_2` = final temperature of gas =
`100^oC=273+100=373K`

Now put all the given values in the above formula, we get the final volume of the gas.

`(88.305L)/(V_2)=(253K)/(373K)`

`V_2=130.188L`

Now we have to calculate the total change of volume.

`V_2-V_1=130.188-88.305=41.883L`

Therefore, the total change of volume is, 41.883 liters.