Solved Examples - Thermodynamics Class 11 Notes | EduRev

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JEE : Solved Examples - Thermodynamics Class 11 Notes | EduRev

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Example 1. Show that pressure of a fixed amount of an ideal gas is a state function  Solved Examples - Thermodynamics Class 11 Notes | EduRev
Solution.
Solved Examples - Thermodynamics Class 11 Notes | EduRev 
Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev = Solved Examples - Thermodynamics Class 11 Notes | EduRevSolved Examples - Thermodynamics Class 11 Notes | EduRev


Example 2. Calculate work done for the expansion of a substance 3m3 to 5m3 against.
(a) Constant pressure = 105 Pa
(b) A veriable pressure = (10 + 5V) Pa

Solution. 
(a) W = -PdV = -105(5-3) = -2 ×105 J

(b)

Solved Examples - Thermodynamics Class 11 Notes | EduRev


Example 3. Calculate change in internal energy for a gas under going from state-I 
(300 K, 2 ×10-2 m3) to state -II (400 K, 4 ×10-2 m3) for one mol. of vanderwaal gas. 
(a) If gas is ideal [Cv = 12 J/K/mol] 
(b) If gas is real 

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Cv = 12 J/k/ mol
a = 2 J.m./mol2

Solution. 
(a) DU = nCV(T2 - T1)
= 1 × 12 ×100 = 1200 J
(b)
Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev


Example 4. One mole of an ideal gas is expanded isothermally at 300 K from 10 atm to 1 atm. Calculate q, w, DU & DH under the following conditions. 
(i) Expansion is carried out reversibly.
(ii) Expansion is carried out irreversibly 
Solution. 
Isothermal process
(i) For ideal gas ΔU = 0, ΔH = 0
q = - w

Solved Examples - Thermodynamics Class 11 Notes | EduRev

(ii)

Solved Examples - Thermodynamics Class 11 Notes | EduRev 

Example 5. Calculate work done for an ideal gas (ln 2 = 0.7)

Solved Examples - Thermodynamics Class 11 Notes | EduRevSolution.
w = -nRT ln
Solved Examples - Thermodynamics Class 11 Notes | EduRev
= 22.4 × 0.7 = 15.68


Example 6. Calculate w = ? (ln 2 = 0.7) 

Solved Examples - Thermodynamics Class 11 Notes | EduRevSolution. 
P - V Relation from plot
y = mx + C
1 = m + C .............(1)
4 = 2m + C .............(2)
⇒ m = 3, C = -2

Hence,
Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev


Example 7.  One mole an ideal gas is expanded from (10 atm, 10 lit) to (2 atm, 50 litre) isothermally. First against 5 atm then against 2 atm. Calculate work done in each step and compare it with single step work done with 2 atm. 
Solution. 
(i) P1V1 = P2V

Solved Examples - Thermodynamics Class 11 Notes | EduRevwirrev = - Pext(V2 - V1) = - 5 (20 - 10) = - 50 atm lit.
(ii) Work done against 2 atm
P1 V1Solved Examples - Thermodynamics Class 11 Notes | EduRev P2 V2
5 atm 20 lit 2 atm 50 lit
wirrev = - Pext(V2 - V1) = - 2 (50 - 20) = - 60 atm lit.
wtotal = - 50 - 60 = - 110 atm lit.
P1 V1Solved Examples - Thermodynamics Class 11 Notes | EduRev P2 V2
10 atm 10 lit 2 atm 50 lit
w = - 2 (50 - 10) = - 80 lit
magnitude of work done in more than one step is more than single step work done.


Example 8. For 1 mole of monoatomic gas. Calculate w, DU, DH, q

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solution
Isochoric process
w = 0
q = dU = Cv (T2 - T1)
Solved Examples - Thermodynamics Class 11 Notes | EduRev
DH = CpΔT = R (400 - 300) = 250 R


Example 9. One mole of an non linear triatomic ideal gas is expanded adiabatically at 300 K from 16 atm to 1 atm. 
Calculate Work done under the following conditions.
(i) Expansion is carried out reversibly. 
(ii) Expansion is carried out irreversibly 
Solution
q = 0
w = DU = Cv(T2 - T1)
Cv for triatomic non linear gas = 3R
(i) For rev. process.
Solved Examples - Thermodynamics Class 11 Notes | EduRev , r = 4/3
Solved Examples - Thermodynamics Class 11 Notes | EduRev
ΔU = w = 3R (150 - 300) = - 450 R
(ii) n = 1
- Pext (V2 - V1) = Cv (T2 - T1)

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

-16T2 T1 = 48 T2 - 48 T1
49 T1 = 64 T2
Solved Examples - Thermodynamics Class 11 Notes | EduRev
T2 = 229.69
wirr = Cv(T2 - T1) 3R (229.69 - 300)
= - 210.93 R


Example 10. One mole of an non linear triatomic ideal gas is compressed adiabatically at 300 K from 1 atom to 16 atm. 
Calculate Work done under the following conditions. 
(i) Expansion is carried out reversibly. (ii) Expansion is carried out irreversibly. 
Solution
q = 0 Adiabatic process
(i)
wrev = DU = Cv(T2 - T1)

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

w = 3R (600 - 300) = 900 R

(ii)
Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

- (T2 - 16 T1) = 3R (T2 - T1)
- T2 16 T1 = 3T2 - 3T1
4T2 = 19 T1
Solved Examples - Thermodynamics Class 11 Notes | EduRev
w = ΔU = 3R (1425 - 300) = 3375 R


Example 11. Calculate work done in process BC for 1 mol of an ideal gas if total 600 cal heat is released by the gas in whole process.

Solved Examples - Thermodynamics Class 11 Notes | EduRev
Solution
For cyclic process
dU = 0
q = - w
PV=1× R × 300
PV=1× R × 400
⇒ P(V- V1) = R (400 - 300)
q = - (wBA + wAC + wCB)
- 600 = - (0 + (-) 2) × (400 - 300) + wCB
wCB = 800 cal


Example 12. Calculate entropy change in each step for an ideal gas (monoatomic)

Solved Examples - Thermodynamics Class 11 Notes | EduRev
Solution
Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev


Example 13. One mole of an ideal gas is expanded isothermally at 300 K from 10 atm to 1 atm. Find the values of 
ΔSsys, ΔSsurrΔStotal  under the following conditions. 
(i) Expansion is carried out reversibly. 
(ii) Expansion is carried out irreversibly 
(iii) Expansion is free. 
Solution 
(i) 
Solved Examples - Thermodynamics Class 11 Notes | EduRev
Solved Examples - Thermodynamics Class 11 Notes | EduRev
Solved Examples - Thermodynamics Class 11 Notes | EduRev
(ii)
Solved Examples - Thermodynamics Class 11 Notes | EduRev
Solved Examples - Thermodynamics Class 11 Notes | EduRev
ΔU = 0 = q + w
qirr = pext)(v2 - v1)
Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

(iii) Free expansion ΔT = 0
w = 0
q = 0
Solved Examples - Thermodynamics Class 11 Notes | EduRev
Solved Examples - Thermodynamics Class 11 Notes | EduRev
Solved Examples - Thermodynamics Class 11 Notes | EduRev


Example 14. One mole of an non linear triatomic ideal gas is expanded adiabatically at 300 K from 16 atm to 1 atm. Find the values of DSsys, DSsurr & DStotal under the following conditions. 
(i) Expansion is carried out reversibly. 
(ii) Expansion is carried out irreversibly 
(iii) Expansion is free. 
Solution. 
For non-linear tri-atomic ideal gas
Cv = 3R, Cp = 4R
(i)
Solved Examples - Thermodynamics Class 11 Notes | EduRev
q = 0
ΔSsurr = - ΔSsys = 0
ΔStotal = 0
(ii) First of all we will have to calculate the temperature of the gas after it has undergoes the said adiabatic reversible expansion we have q = 0
ΔU = q + w
nCv (T2 - T1) = - pext (v2 - v1)
Solved Examples - Thermodynamics Class 11 Notes | EduRev

T2 = 229.68 K

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev 
= - 1.068 R + 2.77 R
= 1.702 R
Solved Examples - Thermodynamics Class 11 Notes | EduRev
ΔStotal =ΔSsys = 1.702 R

(iii) In free adiabatic expansion we have
w = 0
q = 0
ΔT = 0
Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev

Solved Examples - Thermodynamics Class 11 Notes | EduRev


Example 15. For the reaction 
N2 + 2O2 → 2NO2 
Given : at 1 atm, 300 K 
SN2 = 180 J /mol/K Cp(N2) = 30 J/mol/K 
SO2 = 220 J /mol/K Cp(O2) = 30 J/mol/K 
SNO2 = 240 J /mol/K Cp(NO2) = 40 J/mol/K  
Calculate 
(i) ΔS300 K, 1 atm 
(ii) ΔS400 K, 1 atm
(iii) ΔS300 K, 5 atm
(iv) ΔS400 K, 5 atm
Solution.
(i) (ΔSr)300 = 2SNO2 - 2SO2 - SN2 
= 2 × 240 - 2 × 220 - 180
= -140 J mol-1k-1

(ΔCp)r = 2Cp(NO2)- 2Cp(O2) - Cp(N2)
= 2 × 40 - 2 × 30 - 30
= -10 J mol-1k-1

(ii) (ΔSr)400 = (ΔSr)300 (ΔCp)r ln
= - 140 - 10 ln (4/3)
= - 142.88 J mol-1k-1

(iii) (ΔSr)300k, 5 atm  
= (ΔSr)300k, 1 atm  ΔngRln
= - 140 (-1)Rln (1/5)
= - 140 R ln 5
=- 140 8.314 ln 5
= - 126.62 J mol-1k-1

(iv) (ΔSr)400k, 5 atm  
= (ΔSr)400k, 1 atm - R ln
= 142.88 R ln 5
=-129.5 J mol-1k-1


Example 16. From the T-S diagram of a reversible carnot engine calculate: 
(i) efficiency 
(ii) workdone per cycle 
(iii) Heat taken from the source and rejected to sink 
(iv) In order to illuminate 10000 bulbs of 40 watt power each calculate the no. of cycle per second the above must go through.
Solution.
(i) 

Solved Examples - Thermodynamics Class 11 Notes | EduRev

(ii) qrev1-2 = ∫Tds = 1000 x 100 J

q3-4 = ∫Tds = - 200 x 100 J

qnet = 800 × 100 = 80 kJ

W = -80 kJ

(iii) 105 J,

2 × 104 J

(iv) 5


Example 17. Calculate entropy change 
H2O(l, 1 atm, 100ºC) → H2O (g, 1 atm, 110ºC) 
H2O(l, 1 atm, 100ºC) → H2O (g, 2 atm, 100ºC) 
ΔHvap  = 40 kJ/mol Cp(l) = 75 J/mol /K Cp(g) = 35 J/mol/K 
Solution. 
For 1 mol
(i) H2O(l, 1 atm, 100ºC) → H2O (g, 1 atm, 100ºC) → H2O(g, 1 atm, 110ºC)
(A)                                              (B)                                    (C)

Solved Examples - Thermodynamics Class 11 Notes | EduRev

(ii) H2O(l, 1 atm, 100ºC) → H2O (g, 1 atm, 100ºC) → H2O(g, 2 atm, 100ºC)

      (A)                                          (B)                                     (C)

Solved Examples - Thermodynamics Class 11 Notes | EduRev


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