2) One kilogram of steam is contained in a horizontal frictionless piston and th

Question

2) One kilogram of steam is contained in a horizontal frictionless piston and the cylinder is heated at constant pressure of 1.013 bar from 125°C to such a temperature that the volume doubles. If the amount of heat that must be added to accomplish this change is 260 kJ, calculate the final temperature of the steam (assume ideal gas law for now), the work the steam does against its surroundings, and the internal energy changes for the process. Does your value seem too high or too low - if so, why?

Explanation / Answer

Amount of steam = 1 Kg = 1/18 = 0.056 kmol = 56 mol

State 1

P = 1.013 bar

T = 125 oC = 398.15 K

now the process is a constant pressure process

State 2

P = 1.013 bar

we will assume ideal gas law as stated in the problem

hence we can write the below state relation for the above equation

(V1/T1) = (V2/T2)

Now as per the question V2 = 2*V1

(V1/398.15) = (2*V1)/T2

T2 = 2*398.15 = 796.3 K

Now work done can be calculated as

W = P*dV = n*R*dT = 56*8.314* (796.3-298.15) = 231930.6696 Joules = 231.930 KJ

Now internal energy is given by

U = Q - W = 260-231.930 = 28.069 KJ

In order to know whether the value is low or high we will have to compare it witht the real value which can be obtained from steam table

Now we will obtain the Specific internal energy value for both the states and subtract it and than compare it with the ideal value

U = 2733.84- 2544.8 = 189.04 KJ/Kg

For 1 Kg we U = 189.04 KJ

As can be seen, the value of internal energy from ideal case is actually much less than the real case and this mainly due to the assumption of Ideal gas which we have taken to find the work done for this case

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