Thermodynamics Question 5 Follow Link Below for question ple

Thermodynamics

Question 5: Follow Link Below for question please.

http://imgur.com/35uSwH0

Round your answers to two decimal places. 100 kg of R-134a at 500 kPa are contained in a piston-cylinder device whose volume is 4.1168 m The piston is now moved until the volume is one-half its original size. This is done such that the pressure of the R-134a does not change. Determine the final temperature and the change in the total energy of the R-134a.

Solution

first apply ideal gas equation

PV= RT

T= PV /R = 5 X 10⁵ X 4.1168 / 8.314 = 2.47 X 10 ⁵ K

Now charle\' s law

T_² = (V_{2/ V1 ) X T1}

_{= (1/2 ) X 2.47 X 10 ⁵ = 1.235 X 10 ⁵ K}

_{Final energy is calculated by P X ( change in volume}

_{5 X 10 ⁵ X ( 4.1168- 2.0584)= 10.292 X 10 ⁵ J}