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.
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.
(1)
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.
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.
Amit Dhorajiya said:
5 years ago
v1 = 0.03.
v2 = 0.06.
v2-v1 = 0.03.
p = 1 MPa= 1000 KPa = 1000 N/(meter square).
W = PdV = 0.03 * 1000 = 30.
Q = U + W.
U = Q - W = 84 - 30.
U = 54.
v2 = 0.06.
v2-v1 = 0.03.
p = 1 MPa= 1000 KPa = 1000 N/(meter square).
W = PdV = 0.03 * 1000 = 30.
Q = U + W.
U = Q - W = 84 - 30.
U = 54.
(21)
Nikhil said:
1 year ago
v1 = 0.03.
v2 = 0.06.
v2-v1 = 0.03.
p = 1 MPa= 1000 KPa = 1000 N/(meter square).
W = PdV.
= 0.03 * 1000 = 30.
Q = U + W.
U = Q - W = 84 - 30.
U = 54.
v2 = 0.06.
v2-v1 = 0.03.
p = 1 MPa= 1000 KPa = 1000 N/(meter square).
W = PdV.
= 0.03 * 1000 = 30.
Q = U + W.
U = Q - W = 84 - 30.
U = 54.
(1)
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.
Sai Teja said:
8 years ago
Sign convention is opposite for work transfer and heat transfer to the system. Heat absorbed means positive i.e, +84 kJ.
Answer is 54 kJ.
Answer is 54 kJ.
(1)
SUDHEER said:
6 years ago
P1 = P2 = 1000 kPa.
1Q2 = 84 kJ,
1W2 = P1 (V2 " V1),
= 1000 (0.06 " 0.03) = 30 kJ.
1Q2 = 1W2 + 1U2.
U2 " U1= 1Q2 " 1W2 = 84 " 30 = 54 kJ.
1Q2 = 84 kJ,
1W2 = P1 (V2 " V1),
= 1000 (0.06 " 0.03) = 30 kJ.
1Q2 = 1W2 + 1U2.
U2 " U1= 1Q2 " 1W2 = 84 " 30 = 54 kJ.
(3)
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.
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