Only those processes are possible in nature which would give an entropy ____ for the system and the surroundings together.
Answer: B
The entropy of an isolated system can never decrease.
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According to the Boltzmann,
Answer: D
This is how Boltzmann introduced statistical concepts to define disorder.
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1 kg of air at 300 K is mixed with 1 kg air at 400 K in a constant pressure process at 100 kPa and Q = 0. Find the entropy generation in the process.
Answer: A
U2 – U1 + W = U2 – U1 + P(V2 – V1) = H2 – H1 = 0
H2 – H1 = mA(h2 – h1)A + mB(h2 – h1)B = mACp(T2 – TA1) + mBCp(T2 – TB1) = 0
T2 =(mATA1 + mBTB1)/(mA + mB) = (TA1/2) + (TB1/2) = 350 K
1S2 gen = mACp ln(T2/TA1) + mBCp ln(T2/TB1)
= 1 × 1.004 ln [350/300] + 1 × 1.004 ln[350/400] = = 0.15477 – 0.13407 = 0.0207 kJ/K.
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The entropy change of a system between two identifiable equilibrium state is ___ when the intervening process is reversible or change of state is irreversible.
Answer: B
To determine the change in entropy, a known reversible path is made to connect the two end states and integration is performed on this path.
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Which of the following relation is correct?
Answer: D
S=K*lnW where S is the entropy, W is the thermodynamic probability, and K is the Boltzmann constant.
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A piston/cylinder contains 1 kg water at 20°C, 150 kPa. The pressure is linear in volume. Heat is added from 600°C source until the water is at 1 MPa, 500°C. Find the total change in entropy.
Answer: C
v1 = 0.001002 m^3/kg; u1 = 83.94 kJ/kg; s1 = 0.2966 kJ/kg K
v2 = 0.35411 m^3/kg; u2 = 3124.3 kJ/kg; s2 = 7.7621 kJ/kg K
1W2 = ½ (1000 + 150) 1 (0.35411 – 0.001002) = 203 kJ
1Q2 = 1(3124.3 – 83.94) + 203 = 3243.4 kJ
m(s2 – s1) = 1(7.7621 – 0.2968) = 7.4655 kJ/K; 1Q2/T(source) = 3.7146 kJ/K
1S2 gen = m(s2 – s1) − 1Q2/T(SOURCE) = ∆Stotal
= ∆S(H2O) + ∆S(source) = 7.4655 – 3.7146 = 3.751 kJ/K.
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The equation dQ=dE+dW holds good for
Answer: A
This equation holds good for any process and for any system.
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The work done by a closed system in a reversible process is always ___ that done in an irreversible process.
Answer: D
A reversible process always produces maximum work.
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A piston cylinder contains air at 600 kPa, 290 K and a volume of 0.01m^3. A constant pressure process gives 54 kJ of work out. Find the final volume of the air.
Answer: C
W = ∫ P dV = PΔV
ΔV = W/P = 54/600 = 0.09 m^3
V2 = V1 + ΔV = 0.01 + 0.09 = 0.1 m^3.
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The entropy of any closed system can increase in which if the following way?
Answer: C
These two processes increase the entropy of a closed system.
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