Mechanical Engineering - Thermodynamics - Discussion
Discussion Forum : Thermodynamics - Section 2 (Q.No. 33)
33.
In an isothermal process,
Discussion:
21 comments Page 1 of 3.
Kajal Patil said:
4 years ago
During isothermal process,
1) Temperature constant.
2) Enthalpy constant.
3) Internal energy constant.
1) Temperature constant.
2) Enthalpy constant.
3) Internal energy constant.
(1)
Hardik said:
4 years ago
Option A & C is correct.
H= (U2-U1) + P(V2-V1).
Since U2-U1=mR(T2-T1) = 0.
So U2 = UW.
H= (U2-U1) + P(V2-V1).
Since U2-U1=mR(T2-T1) = 0.
So U2 = UW.
Pawan said:
5 years ago
No change in Enthalpy when ideal gas is used.
Varu said:
6 years ago
Because for ideal gases enthalpy is a function of temperature only.
Simha said:
6 years ago
During boiling which is an isothermal process the internal energy of the fluid increases.
Jitendra said:
6 years ago
Only option A is correct, because enthalpy and internal energy may be a function of volume and pressure. Only in special case when we talk about isothermal process for ideal gas then option D is correct.
Joe said:
7 years ago
Isothermal process=Temperature constant. i.e dT=0.
du= CvdT (du=0 since dT=0), so internal energy constant.
dh= CpdT, similarly enthalpy constant.
du= CvdT (du=0 since dT=0), so internal energy constant.
dh= CpdT, similarly enthalpy constant.
Daya said:
7 years ago
I think question should contain ideal gas statement.
Ankit kumar singh said:
7 years ago
According to me, options A is correct.
(1)
Kailash said:
8 years ago
From my view, only for option 1 is correct.
2 and 3 are true only for an ideal gas. If gas is ideal than 2 is correct from joule's low. If 2 is true then 3 is also true because for ideal gas we can write pv=Rt (only for ideal gas). Than h=u+pv.
h=u+rt.
u=c (from joule's low).
2 and 3 are true only for an ideal gas. If gas is ideal than 2 is correct from joule's low. If 2 is true then 3 is also true because for ideal gas we can write pv=Rt (only for ideal gas). Than h=u+pv.
h=u+rt.
u=c (from joule's low).
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