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.

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.

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.

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.

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

Q_bf = ΔE_intbf - W_bf = (18 J) – (0 J)= 18 J

Thus the value of Q_bf will be 18 J.

Q_ib = Q_ibf - Q_bf

= (36 J) – (18 J) = 18 J

Therefore the value of Q_ib would be 18 J.

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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 = 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

= p₀ (V₀-4V₀) = -3 p₀V₀

So,   W_abc= W_ab = -3 p₀V₀

From the above observation we conclude that, the work done on the gas along the path abc would be -3 p₀V₀.

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_cV_c – p_bV_b)         (pV = nRT)

= 3/2[(2p₀) (4V₀) -(p₀) (4V₀)]

= 3/2 [8 p₀V₀ - 4 p₀V₀]= 6 p₀V₀

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(2P₀/P₀) 

= 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 p₀V₀ 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.

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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.

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.

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.