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**Example** **1. Show that pressure of a fixed amount of an ideal gas is a state function ****Solution.**

=

**Example ****2. Calculate work done for the expansion of a substance 3m ^{3} to 5m^{3} against.**

(b) A veriable pressure = (10 + 5V) Pa

(a) W = -PdV = -10

(b)

**Example ****3. Calculate change in internal energy for a gas under going from state-I ****(300 K, 2 ×10 ^{-2} m^{3}) to state -II (400 K, 4 ×10^{-2} m^{3}) for one mol. of vanderwaal gas. **

C_{v} = 12 J/k/ mol

a = 2 J.m./mol^{2}

**Solution. **

(a) DU = nC_{V}(T_{2} - T_{1})

= 1 × 12 ×100 = 1200 J

(b)

**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

(ii)

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

**Solution.**

w = -nRT ln

= 22.4 × 0.7 = 15.68

**Example ****6. 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 ****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) P_{1}V_{1} = P_{2}V_{2 }

w_{irrev} = - P_{ext}(V_{2} - V_{1}) = - 5 (20 - 10) = - 50 atm lit.

(ii) Work done against 2 atm

P_{1} V_{1} P_{2} V_{2}

5 atm 20 lit 2 atm 50 lit

w_{irrev} = - P_{ext}(V_{2} - V_{1}) = - 2 (50 - 20) = - 60 atm lit.

w_{total} = - 50 - 60 = - 110 atm lit.

P_{1} V_{1} P_{2} V_{2}

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**

**Solution****. **

Isochoric process

w = 0

q = dU = C_{v} (T_{2} - T_{1})

DH = C_{p}ΔT = 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 = C_{v}(T_{2} - T_{1})

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} (V_{2} - V_{1}) = C_{v} (T_{2} - T_{1})

-16T_{2} T_{1} = 48 T_{2} - 48 T_{1}

49 T_{1} = 64 T_{2}

T_{2} = 229.69

w_{irr} = C_{v}(T_{2} - T_{1}) 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)**

w_{rev} = DU = C_{v}(T_{2} - T_{1})

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

**(ii)**

- (T_{2} - 16 T_{1}) = 3R (T_{2} - T_{1})

- T_{2} 16 T_{1} = 3T_{2} - 3T_{1}

4T_{2} = 19 T_{1}

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.**

**Solution****. **

For cyclic process

dU = 0

q = - w

PV_{1 }=1× R × 300

PV_{2 }=1× R × 400

⇒ P(V_{2 }- V_{1})_{ }= R (400 - 300)

q = - (w_{BA} + w_{AC} + w_{CB})

- 600 = - (0 + (-) 2) × (400 - 300) + w_{CB}

w_{CB} = 800 cal

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

**Solution****. **

**Example**** 13. One mole of an ideal gas is expanded isothermally at 300 K from 10 atm to 1 atm. Find the values of **

Δ**S _{sys, }**Δ

ΔU = 0 = q + w

q

(iii) Free expansion ΔT = 0

w = 0

q = 0

**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 DS _{sys}, DS_{surr} & DS_{total} under the following conditions. **

For non-linear tri-atomic ideal gas

C

q = 0

ΔS

ΔS

ΔU = q + w

nC

T_{2} = 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**** ****15.** **For the reaction ****N _{2} + 2O_{2} → 2NO_{2} **

(i) ΔS

(ii) ΔS

(iii) ΔS

(iv) ΔS

= 2 × 240 - 2 × 220 - 180

= -140 J mol

(ΔCp)_{r} = 2Cp(NO2)- 2Cp(O2) - Cp(N_{2})

= 2 × 40 - 2 × 30 - 30

= -10 J mol^{-1}k^{-1}

**(ii)** (ΔS_{r})_{400} = (ΔS_{r})_{300} (ΔCp)_{r} ln

= - 140 - 10 ln (4/3)

= - 142.88 J mol^{-1}k^{-1}

**(iii)** (ΔS_{r})_{300k, 5 atm }

= (ΔS_{r})_{300k, 1 atm } Δn_{g}Rln

= - 140 (-1)Rln (1/5)

= - 140 R ln 5

=- 140 8.314 ln 5

= - 126.62 J mol^{-1}k^{-1}

**(iv)** (ΔS_{r})_{400k, 5 atm }

= (ΔS_{r})_{400k, 1 atm }- R ln

= 142.88 R ln 5

=-129.5 J mol^{-1}k^{-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)

(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)** 10^{5} J,

2 × 10^{4} J

(iv) 5

**Example** **17. Calculate entropy change ****H _{2}O(l, 1 atm, 100ºC) → H_{2}O (g, 1 atm, 110ºC) **

Δ

For 1 mol

(i) H

(A) (B) (C)

(ii) H_{2}O(l, 1 atm, 100ºC) → H_{2}O (g, 1 atm, 100ºC) → H_{2}O(g, 2 atm, 100ºC)

(A) (B) (C)

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