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Solved Examples on Thermodynamics Question:1 When a system is taken from state i to state f along the path iaf in below figure, it is found that Q = 50 J and W = -20 J. Along the path ibf, Q = 36 J. (a) What is W along the path ibf? (b) If W = +13 J for the curved return path fi, what is Q for this path? (c) Take E_{int,i} = 10 J. What is E_{int.f}? (d) If E_{int,b} = 22 J, find Q for process ib and process bf. Concept: For a thermodynamic system, in which internal energy is the only type of energy the system may have, the law of conservation of energy may be expressed as Q + W = ΔE_{int} So, W= ΔE_{int }-Q And Q= ΔE_{int }– W Here Q is the energy transferred (as heat) between the system and environment, W is the work done on (or by) the system and ΔE_{int} is the change in the internal energy of the system. By convention we have chosen Q to be positive when heat is transferred into the system and W to be positive when work is done on the system. Solution: A system is taken from i to state f along the path iaf, which as shown in below figure. (a) ΔE_{int} along any path between these two points i and f is, ΔE_{int} = Q+W = (50 J)+(-20 J) =30 J To obtain the W along the path ibf, substitute 30 J for ΔE_{int}, 36 J for Q in the equation W= ΔE_{int }–Q, we get, W= ΔE_{int }–Q = (30 J)-(36 J) = -6 J Thus, the work done W along the path ibf would be -6 J. (b) To obtain the Q for curved return path fi, substitute -30 J for ΔE_{int} and +13 J for W in the equation Q= ΔE_{int }– W, we get, Q= ΔE_{int }– W =(-30 J)-(+13 J) = - 43 J Thus the Q for curved return path fi will be -43 J. (c) As, ΔE_{int} = E_{int,f} - E_{int,i} So, E_{int,f} = ΔE_{int} + E_{int,i} To obtain E_{int,f}, substitute 10 J for E_{int,i} and 30 J for ΔE_{int} in the equation E_{int,f} = ΔE_{int} + E_{int,i} we get, E_{int,f} = ΔE_{int} + E_{int,i} = (30 J)+(10 J) = 40 J Therefore, the value of E_{int,f} would be 40 J. (d) ΔE_{intib} = (22 J) – (10 J) =12 J But, ΔE_{intbf} = (40 J) - (22 J) =18 J And there is no work done on the path bf, since volume is constant. So, Q_{bf} = ΔE_{intbf} - W_{bf} And Q_{ib }= Q_{ibf} - Q_{bf} To obtain the value of Q_{bf}, substitute 18 J for ΔE_{intbf} and 0 J for W in the equation Q_{bf} = ΔE_{intbf} - W_{bf}, we get, Q_{bf} = ΔE_{intbf} - W_{bf }= (18 J) – (0 J)= 18 J Thus the value of Q_{bf} will be 18 J. To obtain the value of Q_{ib}, substitute 36 J for Q_{ibf} and 18 J for Q_{bf} in the equation Q_{ib }= Q_{ibf} - Q_{bf}, we get, Q_{ib }= Q_{ibf} - Q_{bf} = (36 J) – (18 J) = 18 J Therefore the value of Q_{ib} would be 18 J. _____________________________________________________________________________________ Question:2 One mole of an ideal monoatomic gas is caused to go through the cycle shown in below figure. (a) How much work is done on the gas in expanding the gas from a to c along the path abc? (b) What is the change in internal energy and entropy in going through one complete cycle? Express all answers in terms of the pressure p_{0} and volume V_{0} at point a in the diagram. Concept: Work done W at constant pressure is defined as, W = pΔV Here p is the pressure and ΔV is the change in volume. Change in internal energy ΔE is defined as, ΔE = 3/2 nRΔT Here n is the number of moles, R is the gas constant and ΔT is the change in temperature. Change in entropy ΔS is defined as, ΔS = 3/2 ln (T_{f}/T_{i}) Here, T_{f} is the final temperature and T_{i} is the initial temperature. One mole of an ideal monoatomic gas is caused to go through the cycle as shown in the above figure. Solution: (a) To obtain the work done W_{abc} along the path abc, first we have to find out the work done W_{ab} along the path ab. From the figure we observed that, W_{ab} = pΔV = p_{0} (V_{0}-4V_{0}) = -3 p_{0}V_{0} So, W_{abc}= W_{ab} = -3 p_{0}V_{0} From the above observation we conclude that, the work done on the gas along the path abc would be -3 p_{0}V_{0}. (b) The change in internal energy ΔE_{bc} along the path b to c would be, ΔE_{bc} = 3/2 nRΔT_{bc} = 3/2 (nRT_{c} – nRT_{b}) = 3/2(p_{c}V_{c }– p_{b}V_{b}) (pV = nRT) = 3/2[(2p_{0}) (4V_{0}) -(p_{0}) (4V_{0})] = 3/2 [8 p_{0}V_{0} - 4 p_{0}V_{0}]= 6 p_{0}V_{0} The change in entropy ΔS_{bc} along the path b to c would be, ΔS_{bc} = 3/2nR ln(T_{c}/T_{b}) = 3/2nR ln(P_{c}/P_{b}) (Since, T_{c}/T_{b} = P_{c}/P_{b}) =3/2nR ln(2P_{0}/P_{0}) = 3/2 nR ln2 From the above observation we conclude that, the change in internal energy ΔE_{bc} in the path b to c would be 6 p_{0}V_{0} and the change in entropy ΔS_{bc} in the path b to c would be 3/2 nR ln2. (c) The change in entropy and internal energy in a cyclic process is zero. Since the process is a complete cycle, therefore the change in entropy and internal energy in going through one complete cycle would be zero. ___________________________________________________________________________________________ Question:3 For the Carnot cycle shown in below figure, calculate (a) the heat that enters and (b) the work done on the system. Concept: In an isothermal process, the heat Q transfer takes place at constant temperature T. The entropy ΔS of the system is defined as, ΔS = Q/T Here Q is the positive heat flow into the system. So the entropy ΔS of the system will increase. From the equation ΔS = Q/T, the absorbed heat Q will be, Q = (ΔS) (T) = T (S_{f} – S_{i}) Here S_{f} is the final entropy of the system and S_{i} is the initial entropy of he system. For a cyclic path, Q + W = 0 Here Q is the heat added to the system and W is the work done. Solution: The above figure shows a Carnot cycle: (a) We have to calculate the heat Q_{in} that enters into the system. In the above cycle, the heat only enters along the top path AB. To obtain the heat Q_{in} that enters into the system, substitute 400 K for T, 0.6 J/K for S_{f} and 0.1 J/K for S_{i} in the equation Q = T (S_{f} – S_{i}), Q_{in} = T (S_{f} – S_{i}) = (400 K) (0.6 J/K - 0.1 J/K) = (400 K) (0.5 J/K) = 200 J Therefore the heat Q_{in} that enters into the system would be 200 J. (b) To find the work done, first we have to find out the heat that leaves Q_{out} from the bottom path of the cycle. In the above cycle, the heat only leaves along the bottom path CD. Q_{out} = T (S_{f} – S_{i}) = (250 K) (0.1 J/K - 0.6 J/K) = (250 K) (-0.5 J/K) = -125 J Thus the heat that leaves Q_{out} from the bottom path of the cycle would be -125 J. As, Q + W = 0 for a cyclic path, so work done W will be, W = -Q = - (Q_{in}+ Q_{out}) (Since, Q = Q_{in}+ Q_{out}) To obtain the work done W on the system, substitute 200 J for Q_{in} and -125 J for Q_{out} in the equation W = - (Q_{in}+ Q_{out}), W = - (Q_{in}+ Q_{out}) = - (200 J+(-125 J) ) = -75 J From the above observation we conclude that, the work done W on the system would be -75 J.
When a system is taken from state i to state f along the path iaf in below figure, it is found that Q = 50 J and W = -20 J. Along the path ibf, Q = 36 J. (a) What is W along the path ibf? (b) If W = +13 J for the curved return path fi, what is Q for this path? (c) Take E_{int,i} = 10 J. What is E_{int.f}? (d) If E_{int,b} = 22 J, find Q for process ib and process bf.
For a thermodynamic system, in which internal energy is the only type of energy the system may have, the law of conservation of energy may be expressed as
Q + W = ΔE_{int}
So, W= ΔE_{int }-Q
And
Q= ΔE_{int }– W
Here Q is the energy transferred (as heat) between the system and environment, W is the work done on (or by) the system and ΔE_{int} is the change in the internal energy of the system.
By convention we have chosen Q to be positive when heat is transferred into the system and W to be positive when work is done on the system.
A system is taken from i to state f along the path iaf, which as shown in below figure.
(a) ΔE_{int} along any path between these two points i and f is,
ΔE_{int} = Q+W
= (50 J)+(-20 J)
=30 J
To obtain the W along the path ibf, substitute 30 J for ΔE_{int}, 36 J for Q in the equation W= ΔE_{int }–Q, we get,
W= ΔE_{int }–Q
= (30 J)-(36 J) = -6 J
Thus, the work done W along the path ibf would be -6 J.
(b)
To obtain the Q for curved return path fi, substitute -30 J for ΔE_{int} and +13 J for W in the equation Q= ΔE_{int }– W, we get,
=(-30 J)-(+13 J) = - 43 J
Thus the Q for curved return path fi will be -43 J.
(c)
As, ΔE_{int} = E_{int,f} - E_{int,i}
So, E_{int,f} = ΔE_{int} + E_{int,i}
To obtain E_{int,f}, substitute 10 J for E_{int,i} and 30 J for ΔE_{int} in the equation E_{int,f} = ΔE_{int} + E_{int,i} we get,
E_{int,f} = ΔE_{int} + E_{int,i}
= (30 J)+(10 J) = 40 J
Therefore, the value of E_{int,f} would be 40 J.
(d)
ΔE_{intib} = (22 J) – (10 J) =12 J
But,
ΔE_{intbf} = (40 J) - (22 J) =18 J
And there is no work done on the path bf, since volume is constant.
So, Q_{bf} = ΔE_{intbf} - W_{bf}
And Q_{ib }= Q_{ibf} - Q_{bf}
To obtain the value of Q_{bf}, substitute 18 J for ΔE_{intbf} and 0 J for W in the equation Q_{bf} = ΔE_{intbf} - W_{bf}, we get,
Q_{bf} = ΔE_{intbf} - W_{bf }= (18 J) – (0 J)= 18 J
Thus the value of Q_{bf} will be 18 J.
To obtain the value of Q_{ib}, substitute 36 J for Q_{ibf} and 18 J for Q_{bf} in the equation Q_{ib }= Q_{ibf} - Q_{bf}, we get,
Q_{ib }= Q_{ibf} - Q_{bf}
= (36 J) – (18 J) = 18 J
Therefore the value of Q_{ib} would be 18 J.
_____________________________________________________________________________________
One mole of an ideal monoatomic gas is caused to go through the cycle shown in below figure. (a) How much work is done on the gas in expanding the gas from a to c along the path abc? (b) What is the change in internal energy and entropy in going through one complete cycle? Express all answers in terms of the pressure p_{0} and volume V_{0} at point a in the diagram.
Work done W at constant pressure is defined as,
W = pΔV
Here p is the pressure and ΔV is the change in volume.
Change in internal energy ΔE is defined as,
ΔE = 3/2 nRΔT
Here n is the number of moles, R is the gas constant and ΔT is the change in temperature.
Change in entropy ΔS is defined as,
ΔS = 3/2 ln (T_{f}/T_{i})
Here, T_{f} is the final temperature and T_{i} is the initial temperature.
One mole of an ideal monoatomic gas is caused to go through the cycle as shown in the above figure.
(a)
To obtain the work done W_{abc} along the path abc, first we have to find out the work done W_{ab} along the path ab.
From the figure we observed that,
W_{ab} = pΔV
= p_{0} (V_{0}-4V_{0}) = -3 p_{0}V_{0}
So, W_{abc}= W_{ab} = -3 p_{0}V_{0}
From the above observation we conclude that, the work done on the gas along the path abc would be -3 p_{0}V_{0}.
The change in internal energy ΔE_{bc} along the path b to c would be,
ΔE_{bc} = 3/2 nRΔT_{bc}
= 3/2 (nRT_{c} – nRT_{b})
= 3/2(p_{c}V_{c }– p_{b}V_{b}) (pV = nRT)
= 3/2[(2p_{0}) (4V_{0}) -(p_{0}) (4V_{0})]
= 3/2 [8 p_{0}V_{0} - 4 p_{0}V_{0}]= 6 p_{0}V_{0}
The change in entropy ΔS_{bc} along the path b to c would be,
ΔS_{bc} = 3/2nR ln(T_{c}/T_{b})
= 3/2nR ln(P_{c}/P_{b}) (Since, T_{c}/T_{b} = P_{c}/P_{b})
=3/2nR ln(2P_{0}/P_{0})
= 3/2 nR ln2
From the above observation we conclude that, the change in internal energy ΔE_{bc} in the path b to c would be 6 p_{0}V_{0} and the change in entropy ΔS_{bc} in the path b to c would be 3/2 nR ln2.
The change in entropy and internal energy in a cyclic process is zero. Since the process is a complete cycle, therefore the change in entropy and internal energy in going through one complete cycle would be zero.
___________________________________________________________________________________________
For the Carnot cycle shown in below figure, calculate (a) the heat that enters and (b) the work done on the system.
In an isothermal process, the heat Q transfer takes place at constant temperature T. The entropy ΔS of the system is defined as,
ΔS = Q/T
Here Q is the positive heat flow into the system. So the entropy ΔS of the system will increase.
From the equation ΔS = Q/T, the absorbed heat Q will be,
Q = (ΔS) (T)
= T (S_{f} – S_{i})
Here S_{f} is the final entropy of the system and S_{i} is the initial entropy of he system.
For a cyclic path,
Q + W = 0
Here Q is the heat added to the system and W is the work done.
The above figure shows a Carnot cycle:
(a) We have to calculate the heat Q_{in} that enters into the system.
In the above cycle, the heat only enters along the top path AB.
To obtain the heat Q_{in} that enters into the system, substitute 400 K for T, 0.6 J/K for S_{f} and 0.1 J/K for S_{i} in the equation Q = T (S_{f} – S_{i}),
Q_{in} = T (S_{f} – S_{i})
= (400 K) (0.6 J/K - 0.1 J/K)
= (400 K) (0.5 J/K)
= 200 J
Therefore the heat Q_{in} that enters into the system would be 200 J.
(b) To find the work done, first we have to find out the heat that leaves Q_{out} from the bottom path of the cycle.
In the above cycle, the heat only leaves along the bottom path CD.
Q_{out} = T (S_{f} – S_{i})
= (250 K) (0.1 J/K - 0.6 J/K)
= (250 K) (-0.5 J/K)
= -125 J
Thus the heat that leaves Q_{out} from the bottom path of the cycle would be -125 J.
As, Q + W = 0 for a cyclic path, so work done W will be,
W = -Q
= - (Q_{in}+ Q_{out}) (Since, Q = Q_{in}+ Q_{out})
To obtain the work done W on the system, substitute 200 J for Q_{in} and -125 J for Q_{out} in the equation W = - (Q_{in}+ Q_{out}),
W = - (Q_{in}+ Q_{out}) = - (200 J+(-125 J) ) = -75 J
From the above observation we conclude that, the work done W on the system would be -75 J.
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