Mechanical Engineering - Thermodynamics - Discussion
Discussion Forum : Thermodynamics - Section 2 (Q.No. 39)
39.
A mixture of gas expands from 0.03 m3 to 0.06 m3 at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. The change in internal energy of the mixture is
Discussion:
19 comments Page 2 of 2.
ABHISHEK(IIT DHANBAD) said:
8 years ago
Heat is added to the gas so +84kj (heat is given to the system),
pdv work is 1m * (.06-.03) = 30kj,
dQ =dU +pdv,
Gives du = 54.
pdv work is 1m * (.06-.03) = 30kj,
dQ =dU +pdv,
Gives du = 54.
(1)
Suresh kumar said:
9 years ago
@Vdp.
Yes, you are correct according to thermodynamic rules it should be in negative and answer value should be 114 KJ.
Yes, you are correct according to thermodynamic rules it should be in negative and answer value should be 114 KJ.
Nowshad Suhail said:
7 years ago
As the heat is given to system, so we take heat given as negative which ultimately leads us to 114.
Amar Bangar said:
1 decade ago
Change in internal energy
Q=dE+w
84 = dE + 1*10^3*(.06-.03)
dE = 84 - 30
dE = 54 kJ.
Q=dE+w
84 = dE + 1*10^3*(.06-.03)
dE = 84 - 30
dE = 54 kJ.
Anandan said:
9 years ago
@Sandeep.
Please explain. Why sign convention is not applied to nonflow processs?
Please explain. Why sign convention is not applied to nonflow processs?
Sandeep said:
9 years ago
Absolutely wrong @Vdp.
It's a nonflow process. 54 kj is the answer.
It's a nonflow process. 54 kj is the answer.
PKG said:
1 decade ago
dU=dQ-dW.
= [(84*10^3)-{1*10^5*(.06-.03)}] J.
= 81000 J.
= 81 KJ.
= [(84*10^3)-{1*10^5*(.06-.03)}] J.
= 81000 J.
= 81 KJ.
Abdush shami said:
9 years ago
Yes, it should be 114 KJ.
GANGADHAR said:
8 years ago
The correct Answer=54kj.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers