Solved Examples for JEE: Thermodynamics | Chemistry for JEE Main & Advanced PDF Download

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Example 1. In which of the paths between initial state i and final state f in the below figure is the work done on the gas the greatest?

(A) A
(B) B
(C) C
(D) D

Solution. The correct option is (D).

Work is a path function. So work done on the gas depends upon the path. The area under pv diagram gives the work done on the gas between initial state i and final state f. As the area under the path D is greater than the other path, therefore the paths between initial state i and final state f where the work done on the gas is greatest would be D. From the above observation we conclude that, option D is correct. 

Example 2. A 1-kg block of ice at 0ºC is placed into a perfectly insulated, sealed container that has 2 kg of water also at 0ºC. The water and ice completely fill the container, but the container is flexible. After some time one can except that
(A) the water will freeze so that the mass of the ice will increase.
(B) the ice will melt so that the mass of the ice will decrease.
(C) both the amount of water and the amount of ice will remain constant.
(D) both the amount of water and the amount of ice will decrease.

Solution: The correct option is (B) the ice will melt so that the mass of the ice will decrease.
In accordance to second law of thermodynamics, if an irreversible process occurs in a closed system, the entropy of that system always increases; it never decreases. Entropy measures the state of disorder. As the ice is more order than water, therefore the ice will melt so that the mass of the ice will decrease. From the above observation we conclude that, option (B) is correct.  

Example 3. Which type of ideal gas will have the largest value for C_p – C_v?
(A) Monoatomic
(B) Diatomic
(C) Polyatomic
(D) The value will be the same for all.
Solution: 
The correct option is (D).
The specific heat at constant volume (C_v) for monoatomic gas is,
C_v = 3/2 R
The specific heat at constant volume (C_v) for diatomic gas is,
C_v = 5/2 R
The specific heat at constant volume (C_v) for polyatomic gas is,
C_v = 3 R
But we know that, for an ideal gas,
C_p - C_v = R
So,the specific heat at constant pressure (C_p) for monoatomic gas will be,
C_p = C_v + R = 3/2 R + R = 5/2 R
the specific heat at constant pressure (C_p) for diatomic gas will be,
C_p = C_v + R = 5/2 R + R = 7/2 R
the specific heat at constant pressure (C_p) for polyatomic gas will be,
C_p = C_v + R = 3 R + R = 4 R
Therefore,
C_p - C_v for monoatomic gas will be,
C_p - C_v = 5/2 R - 3/2 R = R 
C_p - C_v fo diatomic gas will be,
C_p - C_v = 7/2 R - 5/2 R = R 
C_p - C_v fo polyatomic gas will be,
C_p - C_v = 4 R - 3 R = R 
From the above observation we conclude that, the value of C_p - C_v will be same for all ideal gas. Therefore option (D) is correct.

Example 4. Show that pressure of a fixed amount of an ideal gas is a state function  
Solution.

 =

Example 5. 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)

Example 6. 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 [C_v = 12 J/K/mol] 
(b) If gas is real 

C_v = 12 J/k/ mol
a = 2 J.m./mol²

Solution. 
(a) DU = nC_V(T₂ - T₁)
= 1 × 12 ×100 = 1200 J
(b)

Example 7. 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

(ii)

Example 8. Calculate work done for an ideal gas (ln 2 = 0.7)

Solution.
w = -nRT ln

= 22.4 × 0.7 = 15.68

Example 9. Calculate w = ? (ln 2 = 0.7) 

Solution. 
P - V Relation from plot
y = mx + C
1 = m + C .............(1)
4 = 2m + C .............(2)
⇒ m = 3, C = -2

Hence,

Example 10.  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 = P₂V₂ 

w_irrev = - P_ext(V₂ - V₁) = - 5 (20 - 10) = - 50 atm lit.
(ii) Work done against 2 atm
P₁ V₁ P₂ V₂
5 atm 20 lit 2 atm 50 lit
w_irrev = - P_ext(V₂ - V₁) = - 2 (50 - 20) = - 60 atm lit.
w_total = - 50 - 60 = - 110 atm lit.
P₁ V₁ 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 11. For 1 mole of monoatomic gas. Calculate w, DU, DH, q

Solution. 
Isochoric process
w = 0
q = dU = C_v (T₂ - T₁)

DH = C_pΔT = R (400 - 300) = 250 R

Example 12. 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 = C_v(T₂ - T₁)
C_v for triatomic non linear gas = 3R
(i) For rev. process.
, r = 4/3

ΔU = w = 3R (150 - 300) = - 450 R
(ii) n = 1
- P_ext (V₂ - V₁) = C_v (T₂ - T₁)

-16T₂ T₁ = 48 T₂ - 48 T₁
49 T₁ = 64 T₂

T₂ = 229.69
w_irr = C_v(T₂ - T₁) 3R (229.69 - 300)
= - 210.93 R

Example 13. 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)
w_rev = DU = C_v(T₂ - T₁)

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

(ii)

- (T₂ - 16 T₁) = 3R (T₂ - T₁)
- T₂ 16 T₁ = 3T₂ - 3T₁
4T₂ = 19 T₁

w = ΔU = 3R (1425 - 300) = 3375 R

Example 14. 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.

Solution. 
For cyclic process
dU = 0
q = - w
PV₁ =1× R × 300
PV₂ =1× R × 400
⇒ P(V₂ - V₁) = R (400 - 300)
q = - (w_BA + w_AC + w_CB)
- 600 = - (0 + (-) 2) × (400 - 300) + w_CB
w_CB = 800 cal

Example 15. Calculate entropy change in each step for an ideal gas (monoatomic)

Solution. 

Example 16. One mole of an ideal gas is expanded isothermally at 300 K from 10 atm to 1 atm. Find the values of 
ΔS_sys, ΔS_surr & ΔS_total  under the following conditions. 
(i) Expansion is carried out reversibly. 
(ii) Expansion is carried out irreversibly 
(iii) Expansion is free. 
Solution.  
(i) 



(ii)


ΔU = 0 = q + w
q_irr = p_ext)(v₂ - v₁)

(iii) Free expansion ΔT = 0
w = 0
q = 0

Example 17. 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 DS_sys, DS_surr & DS_total 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
C_v = 3R, C_p = 4R
(i)

q = 0
ΔS_surr = - ΔS_sys = 0
ΔS_total = 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
nC_v (T₂ - T₁) = - p_ext (v₂ - v₁)

T₂ = 229.68 K

= - 1.068 R + 2.77 R
= 1.702 R

ΔS_total =ΔS_sys = 1.702 R

(iii) In free adiabatic expansion we have
w = 0
q = 0
ΔT = 0

Example 18. For the reaction 
N₂ + 2O₂ → 2NO₂ 
Given : at 1 atm, 300 K 
S_N2 = 180 J /mol/K C_p(N₂) = 30 J/mol/K 
S_O2 = 220 J /mol/K C_p(O₂) = 30 J/mol/K 
S_NO2 = 240 J /mol/K C_p(NO₂) = 40 J/mol/K  
Calculate 
(i) ΔS_{300 K, 1 atm} 
(ii) ΔS_{400 K, 1 atm}
(iii) ΔS_{300 K, 5 atm}
(iv) ΔS_{400 K, 5 atm}
Solution.
(i) (ΔS_r)₃₀₀ = 2S_NO2 - 2S_O2 - S_N2
= 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) (ΔS_r)₄₀₀ = (ΔS_r)₃₀₀ (ΔCp)_r ln
= - 140 - 10 ln (4/3)
= - 142.88 J mol^-1k^-1

(iii) (ΔS_r)_{300k, 5 atm} 
= (ΔS_r)_{300k, 1 atm}  Δn_gRln
= - 140 (-1)Rln (1/5)
= - 140 R ln 5
=- 140 8.314 ln 5
= - 126.62 J mol^-1k^-1

(iv) (ΔS_r)_{400k, 5 atm} 
= (ΔS_r)_{400k, 1 atm} - R ln
= 142.88 R ln 5
=-129.5 J mol^-1k^-1

Example 19. 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) 

(ii) q_rev1-2 = ∫Tds = 1000 x 100 J

q_3-4 = ∫Tds = - 200 x 100 J

q_net = 800 × 100 = 80 kJ

W = -80 kJ

(iii) 10⁵ J,

2 × 10⁴ J

(iv) 5

Example 120. Calculate entropy change 
H₂O(l, 1 atm, 100ºC) → H₂O (g, 1 atm, 110ºC) 
H₂O(l, 1 atm, 100ºC) → H₂O (g, 2 atm, 100ºC) 
ΔH_vap  = 40 kJ/mol C_p(l) = 75 J/mol /K C_p(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)

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

      (A)                                          (B)                                     (C)