Thermo Solved Questions
QUESTION 1- Constant Volume Heat Addition
(from www.msubbu.com)
A 28 liter rigid enclosure contains air at 140 kPa and 20oC. Heat is added to the container until the pressure reaches 345 kPa. Calculate the heat added.
SOLUTION:
Given: V = 28L= 0.028m (1L = 1dm3; 10dm = 1m so (10dm)3 = (1m)3)
P1 = 140 KPa
T1 = 20oC = 293K
P2 = 345KPa
∆Q = ?
From the 1st Law of Thermodynamics,
∆Q = ∆U + W
Introducing the reversible work terms,
∆Q = ∆U + P∆V
But at constant volume, P∆V = 0
So ∆Q = ∆U
And ∆Q = ∆U = mCp∆T
But what is m? Cp?
-m can be gotten from the formula PV = nRT, where n = m/M (mass/molecular mass)
m = nM, and n = PV/RT = (140 x 0.028)/(8.314 x 293)
n = 1.609 x 10^-3, and M air = 29, so m = 1.609 x 10^-3 x 29 = 0.0467kg
Good. Now,
-Cv is used, not Cp, because we are dealing wit constant volume here.
Cv air at 20C = 0.718KJ/KgC
Also, at constant V, P ∞ T
So P1/T1 = P2/T2 and T2 = 722K
and ∆Q = ∆U = mCv∆T = 0.0467 x 0.718 x (722 – 293)
∆Q = heat added = 14.385 J//
QUESTION 2: First Law, Cycle Work
A thermodynamic system undergoes a cycle composed of a series of three processes for which Q1 = +10 kJ, Q2 = +30 kJ, Q3 = -5 kJ. For the first process, ∆E = +20 kJ, and for the third process, ∆E = -20 kJ. What is the work in the second process, and the net work output of the cycle?
( www.msubbu.com)
SOLUTION:
Q + W = ∆U
Work done in the first process = 20 - 10 = 10 kJ (i.e., work is done on the system)
Work done in the third process = -20 - (-5) = -15 kJ
For a cyclic process, the overall internal energy change is zero. (i.e., ∆U = 0)
Therefore, ∆U in the second process = (0 - (20 - 20)) = 0 kJ
Therefore, work done in the second process = 0 - 30 kJ = -30 kJ.
Total work done during the cycle = 10 + (-15) + (-30) = -35 kJ (i.e., 35 kJ of work is done by the system).
Total work done can also be calculated as follows:
Total Q = Q1 + Q2 + Q3 = 10 + 30 - 5 = 35 kJ
Therefore, net work done during the cycle = 0 - 35 kJ = -35 kJ (the negative sign indicates that work is done by the system//
QUESTION 3: Constant Internal Energy Process
A tank having a volume of 0.1 m3 contains air at 14 MPa and 50oC. It is connected through a valve to a larger tank having a volume of 15 m3, which is completely evacuated. The entire assembly is completely insulated. The valve is opened and the gas allowed to come to equilibrium in both tanks. Calculate the final pressure.
(From www.msubbu.com)
SOLUTION:
Given:
V1 = 0.1m3
P1 = 14MPa = 14 x 1000KPa
T = 50C = 323K
V2 = 15m3
∆Q = 0 (since the system is insulated, ie. isolated system)
P2 = ?
∆U = ∆Q + W,
and ∆U = ∆Q + P∆V
but ∆Q = 0, and while volume changes, work is not actually done on the system- it is just a transfer of air from one volume to another volume (imagine it to be ‘unaided diffusion’), so W= 0
so ∆U = 0 and ∆T = 0 since ∆U = mCv∆T
Since T is constant, P1V1 = P2V2
14 x 0.1 = P2 x (0.1 + 15)
Thus P2 = 92.715KPa//
(from www.msubbu.com)
A 28 liter rigid enclosure contains air at 140 kPa and 20oC. Heat is added to the container until the pressure reaches 345 kPa. Calculate the heat added.
SOLUTION:
Given: V = 28L= 0.028m (1L = 1dm3; 10dm = 1m so (10dm)3 = (1m)3)
P1 = 140 KPa
T1 = 20oC = 293K
P2 = 345KPa
∆Q = ?
From the 1st Law of Thermodynamics,
∆Q = ∆U + W
Introducing the reversible work terms,
∆Q = ∆U + P∆V
But at constant volume, P∆V = 0
So ∆Q = ∆U
And ∆Q = ∆U = mCp∆T
But what is m? Cp?
-m can be gotten from the formula PV = nRT, where n = m/M (mass/molecular mass)
m = nM, and n = PV/RT = (140 x 0.028)/(8.314 x 293)
n = 1.609 x 10^-3, and M air = 29, so m = 1.609 x 10^-3 x 29 = 0.0467kg
Good. Now,
-Cv is used, not Cp, because we are dealing wit constant volume here.
Cv air at 20C = 0.718KJ/KgC
Also, at constant V, P ∞ T
So P1/T1 = P2/T2 and T2 = 722K
and ∆Q = ∆U = mCv∆T = 0.0467 x 0.718 x (722 – 293)
∆Q = heat added = 14.385 J//
QUESTION 2: First Law, Cycle Work
A thermodynamic system undergoes a cycle composed of a series of three processes for which Q1 = +10 kJ, Q2 = +30 kJ, Q3 = -5 kJ. For the first process, ∆E = +20 kJ, and for the third process, ∆E = -20 kJ. What is the work in the second process, and the net work output of the cycle?
( www.msubbu.com)
SOLUTION:
Q + W = ∆U
Work done in the first process = 20 - 10 = 10 kJ (i.e., work is done on the system)
Work done in the third process = -20 - (-5) = -15 kJ
For a cyclic process, the overall internal energy change is zero. (i.e., ∆U = 0)
Therefore, ∆U in the second process = (0 - (20 - 20)) = 0 kJ
Therefore, work done in the second process = 0 - 30 kJ = -30 kJ.
Total work done during the cycle = 10 + (-15) + (-30) = -35 kJ (i.e., 35 kJ of work is done by the system).
Total work done can also be calculated as follows:
Total Q = Q1 + Q2 + Q3 = 10 + 30 - 5 = 35 kJ
Therefore, net work done during the cycle = 0 - 35 kJ = -35 kJ (the negative sign indicates that work is done by the system//
QUESTION 3: Constant Internal Energy Process
A tank having a volume of 0.1 m3 contains air at 14 MPa and 50oC. It is connected through a valve to a larger tank having a volume of 15 m3, which is completely evacuated. The entire assembly is completely insulated. The valve is opened and the gas allowed to come to equilibrium in both tanks. Calculate the final pressure.
(From www.msubbu.com)
SOLUTION:
Given:
V1 = 0.1m3
P1 = 14MPa = 14 x 1000KPa
T = 50C = 323K
V2 = 15m3
∆Q = 0 (since the system is insulated, ie. isolated system)
P2 = ?
∆U = ∆Q + W,
and ∆U = ∆Q + P∆V
but ∆Q = 0, and while volume changes, work is not actually done on the system- it is just a transfer of air from one volume to another volume (imagine it to be ‘unaided diffusion’), so W= 0
so ∆U = 0 and ∆T = 0 since ∆U = mCv∆T
Since T is constant, P1V1 = P2V2
14 x 0.1 = P2 x (0.1 + 15)
Thus P2 = 92.715KPa//