Mechanical Engineering - Thermodynamics - Discussion
Discussion Forum : Thermodynamics - Section 2 (Q.No. 33)
33.
In an isothermal process,
Discussion:
21 comments Page 2 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).
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).
Bhaa said:
9 years ago
A and C only.
Vinoth said:
8 years ago
It is true for ideal gas.
de, dh depends on temperature only.
de, dh depends on temperature only.
Chintan said:
8 years ago
Isothermal temp remains constant.
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.
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.
Prashant said:
8 years ago
In an isothermal process internal energy is constant but work is done so enthalpy must be change.
Adam said:
9 years ago
I think A is correct.
Aman said:
9 years ago
Option B is not the correct one.
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.
Himanshu said:
9 years ago
B is not the correct one.
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