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**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.**

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 = Î”E

So, W= Î”E

And

Q= Î”E

Here Q is the energy transferred (as heat) between the system and environment, W is the work done on (or by) the system and Î”E

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) Î”Eint along any path between these two points i and f is,

Î”E

= (50 J)+(-20 J)

=30 J

To obtain the W along the path ibf, substitute 30 J for Î”E

Î”E

W= Î”E

= (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 Î”E

Q= Î”E

=(-30 J)-(+13 J) = - 43 J

Thus the Q for curved return path fi will be -43 J.

(c)

As, Î”E

So, E

To obtain E

E

= (30 J)+(10 J) = 40 J

Therefore, the value of Eint,f would be 40 J.

(d)

Î”E

But,

Î”E

And there is no work done on the path b

So, Q

And Q

To obtain the value of Q

Î”E

Q

Thus the value of Q

To obtain the value of Q

Q

Q

= (36 J) â€“ (18 J) = 18 J

Therefore the value of Q

**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 p0 and volume V0 at point a in the diagram.****Concept:**

Work done W at constant pressure is defined as,

W = pÎ”V

Here p is the pressure and Î”V is the change in volume.

Change in internal energy Î”E is defined as,

Î”E = 3/2 nRÎ”T

Here n is the number of moles, R is the gas constant and Î”T is the change in temperature.

Change in entropy Î”S is defined as,

Î”S = 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 Wabc along the path abc, first we have to find out the work done Wab along the path ab.

From the figure we observed that,

W_{ab} = pÎ”V

= 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 Î”Ebc along the path b to c would be,

Î”E_{bc }= 3/2 nRÎ”T_{bc}

= 3/2 (nRT_{c} â€“ nRT_{b})

= 3/2(p_{c}V_{c} â€“ 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 Î”S_{bc }along the path b to c would be,

Î”S_{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 Î”Ebc in the path b to c would be 6 p0V0 and the change in entropy Î”Sbc 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 Î”S of the system is defined as,

Î”S = Q/T

Here Q is the positive heat flow into the system. So the entropy Î”S of the system will increase.

From the equation Î”S = Q/T, the absorbed heat Q will be,

Q = (Î”S) (T)

= T (S_{f} â€“ S_{i})

Here S_{f} is the final entropy of the system and Si 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 Qin 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 Si in the equation Q = T (S_{f}â€“ S_{i}),

Q_{in} = T (S_{f} â€“ 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} â€“ 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 Qout 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 Qin and -125 J for Qout 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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