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**LAWS OF THERMODYNAMICS****Zeroth Law**

Based on thermal equilibrium if A & B, & B & C are in thermal equilibrium then A & C must be in thermal equilibrium.

**First Law of Thermodynamics**

First law of thermodynamics is based on energy conservation

E_{2 }= E_{1 + }q + w (E_{1} is the E_{i})

E_{2 }- E_{1 }= q + w (E_{1} q w is E_{f})

(E_{i} = E_{f})

ΔE = Q+w |

For an isolated system, q = 0, w = 0

ΔU = 0

or U = constant

For cyclic process.

Work done = - F_{ext}.dx

= - P. A dx

dW = -P_{ext}dV |

for expansion → dW = - ve

compression → dW = +ve

dq = CdT

⇒ dq_{v} = C_{v}dT

C_{v} = dq_{v}/dT = q_{v}/ΔT

For an isochoric process dv = 0

dU = q_{V}

⇒ dU = C_{V }dTa

dU = n C_{V}dT

ΔU = nC_{v}(T_{2} - T_{1})

If C_{V} is a function of temperature

We know that

U = f(T, V, P)

consider U = f(T, V)

For isochoric process dv = 0

for 1 mole of gas

For an ideal gas U = f(T) only

du = C_{V}dT

ΔU = nC_{V}ΔT = nC_{V}(T_{2} - T_{1})

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

**ENTHALPY**

H = U PV

dH = dU d(PV)

∫dH = ∫dU + P∫dV

at constant pressure

⇒ ΔH = ΔU + PΔV

From, I^{st} law of thermodynamics at constant pressure.

dU = dq_{p} + dW

dU = dq_{p} - PdV

dq_{p} = dU + PdV

dH = dU + PdV

ΔH = ΔU + PΔV ....(1)

at constant volume

ΔH = Δ U + VΔP ..........(2)

When P & V both changes

ΔH = Δ U + (P_{2}V_{2} - P_{1}V_{1}) ..........(3)

ΔH = Δ U + d(PV)

ΔH = Δ U + d(nRT)

ΔH = Δ U + RTDn_{g}

For an ideal gas expansion or compression

ΔH = ΔU + nRΔT |

ΔH = nC_{V}ΔT + nRΔT

= nΔT[C_{V} + R]

ΔH = nΔT C_{p} |

We know that, H = f(T, P)

At constant pressure

dP = 0

**Calculation of Δn _{g} for any chemical reaction:**

If reaction is 50% completed.

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