The document Work Done Calculation and Adiabatic Expansion and Compression (Reversible and Irreversible) Class 11 Notes | EduRev is a part of the Class 11 Course Chemistry Class 11.

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**WORKDONE CALCULATION**

**(1) Isochoric process: **

V = constant

dV = 0

W = 0 |

dU = dq_{V}

ΔU = q_{V } = nC_{V}ΔT

ΔH = nC_{p}ΔT |

**(2) Isobaric process:**

W = - P_{ext} (V_{2}-V_{1})

Reversible & isobaric process

W = - P (V_{2}-V_{1})

= - nR (T_{2}-T_{1})

Irreversible & isobaric process

P_{1} = P_{2} = P_{ext}

For reversible & irreversible isobaric or isochoric process, workdone is same.

**(3) Isothermal process. **

**(a) Reversible Expansion or compression**

W = -P∫ dv

= -∫P_{gas} dV

In Expansion W = - ve

ΔE = 0

q = -W |

**(b) Single stage irreversible expansion**

W = - P_{ext}(V_{2} - V_{1})

|W_{rev}| > |W_{irr}| (in case of expansion)

**(c) Two Stage irreversible Expansion: **

**Stage I. **P_{ext } = 3 atm, P_{i} = 5 atm**Stage II. **P"_{ext } = 2 atm , P_{f} = 2 atm

Workdone in 2^{nd} stage > Workdone in I^{st} stage

**(d) ** **n- stage expansion **

** ****Compression - (One stage Compression ) **

| W_{irr }| = P_{ext} DV

P_{1} = 1 atm , P_{2} = 5 atm , P_{ext} = 5 atm

** **| W_{irr }| > | W_{rev }| For compression

** **

**Example 1. ****2 moles of an ideal gas initially present in a piston fitted cylinder at 300 K, and 10 atm are allowed to expand against 1 atm but the piston was stopped before it stablished the mechanical equilibrium. If temperature were maintained constant through out the change and system delivers 748.26 J of work, determine the final gas pressure and describe the process on PV diagram. ****Solution.**

W_{irrv} = - 748.26

W_{irr} = - P_{ext }[1/P_{2}- 1/P_{1}]nRT

P_{2} = 4atm

**Example 2. ****1150 Kcal heat is released when following reaction is carried out at constant volume. ****C _{7}H_{16(l)} 11O_{2(g)} → 7CO_{2(g)} 8H_{2}O_{(l)} **

**Solution.** 100 = a + bV

⇒ 100 = a + 3b

Also, 400 = a + 6b

⇒ a = -200, b = 100**Δ**U = 34 + 3(P_{2}V_{2} - P_{1}V_{1})

= 6300 J**Δ**H = **Δ**U + P_{2}V_{2} - P_{1}V_{1}

= 6300 + 2100 = 8400 J

P is a linear function

P_{ext} = (400 + 100)/2 = 250

W = - P_{ext}(dV)

= - 250(6 - 3) = - 750 J

**Example 3. ****4 moles of an ideal gas (C _{v} = 15 J) is subjected to the following process represented on P - T graph. From the given data find out whether the process is isochoric or not ? also calculate q, w, **Δ

**Solution.** PV = nRT

4V = 4R × 400

V = 400 R ........(1)

3V = 4R × 300

⇒ V = 400 R ........(2)

i.e., V is constant

w = 0

ΔU = nC_{V} + ΔT ⇒ 4 ×15 ×100 = 6000 J

ΔH = nC_{P} + ΔT ⇒ n (C_{V }+ R ) ΔT

⇒ 4 ×(15 8.3)×100

⇒ 9320 J

q = ΔU = 6kj |

**Example 4. ****2 mole of a gas at 1 bar and 300 K are compressed at constant temperature by use of a constant pressure of 5 bar. How much work is done on the gas ? ****Solution.**

= 19953.6 J

**Example 5. ****2 moles of an ideal diatomic gas (C _{V } = 5/2 R) at 300 K, 5 atm expanded irreversibly and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm. **

w = C

Given** T _{2} = 270K**

P

q = 0

w =

= -1247.1 J

= 1745.94 J

If process is reversible Pv

P

T = 231 K |

**ADIABATIC IDEAL GAS EXPANSION AND COMPRESSION**

dq = 0

dU = dW

Δ U = W |

W = nC_{V }ΔT = nC_{V} T_{2} - T_{1}) |

For an ideal gas C_{P} - C_{V} = R

**REVERSIBLE ADIABATIC EXPANSION OR COMPRESSION**

nC_{V}dT = - P_{ext} dV

P_{int }= dP = P_{ext}

⇒ T_{2}(V_{2})^{γ - 1} = V_{1}^{γ - 1}T_{1}

TV^{γ - 1} = Constant

**IRREVERSIBLE ADIABATIC EXPANSION OR COMPRESSION**

dU = dW

⇒ nC_{V}(T_{2} - T_{1}) = - P_{ext} dV

= - P_{ext} [V_{2} - V_{1}]

**COMPARISON OF REVERSIBLE ISOTHERMAL AND REVERSIBLE ADIABATIC IDEAL GAS EXPANSION ****(i)**** If final volumes are same**

Isothermal process.

P_{1}v_{1} = P_{iso} V_{2 }

⇒

Adiabatic process.

P_{1}v_{1}^{γ } = P_{adia} V_{2}^{γ }

⇒ P |

**(ii) If final pressures are same**

Isothermal process.

P_{1}V_{1} = P_{2} V_{iso }

.......(1)

P_{1}V_{1}^{γ}_{ } = P_{2}V^{γ }_{adia}

In ideal gas expansion,

| W_{iso }| > | W_{adia }|

Hence

⇒ V |

**Compression **

**(i) If final volumes are same**

For isothermal process

P_{1}V_{1} = P_{iso}V_{2}

..........(1)

Adiabatic process.

P_{1}V_{1}^{γ } = P_{adia}V_{2}^{γ }

..........(2)

P |

**(ii) If final pressures are same**

P_{1}V_{1} = P_{2 }V_{iso} ..........(1)

P_{1}V_{1}^{γ } = P_{2 }V^{γ }_{adia} ..........(2)

⇒

⇒

⇒ V |

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