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.
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.
GANGADHAR said:
8 years ago
The correct Answer=54kj.
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.
Jishnu said:
7 years ago
How to calculate the internal energy and enthalpy of a system having mass 1 kilogram and volume 0.03 M cube and pressure 3 mega pascal?
Please, anyone, tell me.
Please, anyone, tell me.
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.
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.
Uday Vadecha said:
4 years ago
The isobaric expansion process is a heat-absorbing process.
So, from 84 kJ heat addition, 30 kJ heat is expended in the Isobaric expansion process & the remaining 54 kJ heat is saved as internal energy.
So, from 84 kJ heat addition, 30 kJ heat is expended in the Isobaric expansion process & the remaining 54 kJ heat is saved as internal energy.
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.
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.
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