Question

value: 20.00 points 1 out of 3 attempts A piston-cylinder device contains 609 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. Assistance Check My Work View Hint View Question Print Question Help Report a Problem

Answer 1

Solution :-

R134 a = CF3CH2F

Molar mass of CF3CH2F = 102.03 g per mol

Lets first calculate moles of 609 g CF3CH2F

Moles = mass / molar mass

Moles of CF3CH2F = 609 g / 102.03g per mol

                                        = 5.969 mol

Initial temperature = -20 C +273 = 253 K

Pressure = 60 kpa * 1 atm / 101.3 kpa

                 = 0.5923 atm

Lets calculate the initial volume at -20 C temperature

PV= nRT

V= nRT/P

   = (5.969 mol * 0.08206 L atm per mol K * 253 K) /0.5923 atm

   = 209.2 L

Now lets calculate the volume when the temperature is 100 C that is 373 K

V =nRT/P

= (5.969 mol * 0.08206 L atm per mol K * 373 K) /0.5923 atm

   = 308.5 L

Now lets calculate the change in the volume

$\Delta$ V = final volume – initial volume

             = 308.5 L – 209.2 L

      = 99.3 L

Now lets convert liter to m3

99.3 L * (1 m3 / 1000 L) = 0.0993 m3

Therefore the volume change is 0.0993 m³