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 1 of 2.
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.
Lalith kailash said:
1 decade ago
Q = 84kJ, Change in volume dV = 0.06-0.03 , pressure = 1Mpa = 0.03
Q = U+PV.
dQ =dU+P.dV.
84 = dU+ 1*10^6*(0.03).
dU = 84-30*10^3.
dU = 54kJ.
Q = U+PV.
dQ =dU+P.dV.
84 = dU+ 1*10^6*(0.03).
dU = 84-30*10^3.
dU = 54kJ.
Dhage Amol said:
1 decade ago
We know that first law of thermodynamics.
dU = dQ-dW.
dU = 84-[p(V2-V1)].
dU = 84-[10^3(0.06-0.03).
dU = 84-[0.03*10^3].
dU = 84-30.
dU = 54 KJ.
dU = dQ-dW.
dU = 84-[p(V2-V1)].
dU = 84-[10^3(0.06-0.03).
dU = 84-[0.03*10^3].
dU = 84-30.
dU = 54 KJ.
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.
Vdp said:
10 years ago
Cant it be like this?
Heat is absorbed during process. So it should be negative as per sign convention.
Q = -84.
And so dU = 114 KJ.
Heat is absorbed during process. So it should be negative as per sign convention.
Q = -84.
And so dU = 114 KJ.
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.
Abdush shami said:
9 years ago
Yes, it should be 114 KJ.
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.
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?
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)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers