Answer:
Q = E + W = 0.25 + 0.4 = 0.29KJ
W = 0.4KJ
b) 2000KJ
Explanation:
The concept applied here is the first law of thermodynamics ,i.e the equation of the first law.
mathematically from first law,
dE = dQ - dW
where E is internal energy
Where energy may be transfereed as eitherf work (W) or heat (Q)
Work on the other hand can be at constant volume and pressure.
at constant pressure, W = Integral (pdV), with final volume(Vf) as the upper limit and initial volume(Vi) as the lower limit
Work = p(Vf -Vi)
attempting the first question, Vi = 0.1metre cube, Vf = 0.12metre cube, p = 2bar, p(atm) = 1bar, E = U = 0.25KJ
from W = p(Vf -Vi) = 2 x 100000 ( 0.12 - 0.1) = 400N/m = 0.4KJ
To get the heat transfer, from dE = dQ - dW
Q = E + W = 0.25 + 0.4 = 0.29KJ
b) for the second question ; from Pdv = p(Vf -Vi), but the pressure here is in atmosphere
W = 100000 ( 0.12 - 0.1) = 2000KJ
