Please show work. Thank you!
Solution
a)
First law of thermodynamics:
q - w = du
where work done w = p*dv = p*(v2 - v1) = 0 since for constant volume process, v1 = v2.
Therefore, q = du
For ideal gas, du = Cv*dT
so q = Cv*dT
For liquids, du = v dp
so, q = v*dp
b)
First law of thermodynamics:
q - w = du
where work done w = p*dv = p*(v2 - v1)
Therefore, q = du + p*dv
q = (u2 - u1) + p*(v2 - v1)
q = (u2 + p*v2) - (u1 + p*v1)
But enthalpy h = u + pv
So, q = h2 - h1
q = dh
For ideal gas, dh = Cp*dT
so q = Cp*dT
For liquids, dh = Cp*(T2 - T1)
.
Please show work- Thank you!Solutiona) First law of thermodynamics- q.docx
1. Please show work. Thank you!
Solution
a)
First law of thermodynamics:
q - w = du
where work done w = p*dv = p*(v2 - v1) = 0 since for constant volume process, v1 = v2.
Therefore, q = du
For ideal gas, du = Cv*dT
so q = Cv*dT
For liquids, du = v dp
so, q = v*dp
b)
First law of thermodynamics:
2. q - w = du
where work done w = p*dv = p*(v2 - v1)
Therefore, q = du + p*dv
q = (u2 - u1) + p*(v2 - v1)
q = (u2 + p*v2) - (u1 + p*v1)
But enthalpy h = u + pv
So, q = h2 - h1
q = dh
For ideal gas, dh = Cp*dT
so q = Cp*dT
For liquids, dh = Cp*(T2 - T1)
