Mechanical Engineering - Thermodynamics - Discussion

Discussion Forum : Thermodynamics - Section 2 (Q.No. 33)
33.
In an isothermal process,
there is no change in temperature
there is no change in enthalpy
there is no change in internal energy
all of these
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
21 comments Page 1 of 3.

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).

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.

Abhishek Tripathy said:   9 years ago
B is not the correct answer because in PV curve when a phase change occurs from liquid to vapor (in the wet region) the temperature is constant but enthalpy change.

ABHISHEK said:   8 years ago
t=c, therefore no change in temp.
t=c, u=f(t) so du=0 or u=c
h=u+pv, here u=c, and t=c gives pv=c finally every thing in this is cont.

So, the answer is D.

VRC said:   9 years ago
D is the correct answer because in an isothermal process T1 = T2, that's why internal energy, temperature & enthalpy remains constant/unchanged.

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.

Divyank singh said:   8 years ago
H=u+Pv And we know that change in pv for isothermal is zero and internal energy is constant so enthalpy is constant from this.

D is correct.

Kajal Patil said:   4 years ago
During isothermal process,

1) Temperature constant.
2) Enthalpy constant.
3) Internal energy constant.
(1)

Prashant said:   8 years ago
In an isothermal process internal energy is constant but work is done so enthalpy must be change.

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.


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