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.
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)
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)
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)
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)
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)
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)
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?
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.
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