# Top Questions with detailed solutions| Thermodynamics-II|CSIR NET, GATE

## Thermodynamics-II

### Top Questions with detailed solutions| Thermodynamics-II

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### Solve thermodynamics Problems with a detailed solution:

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**Q.9. For a system subjected to only P-V work, entropy is given by**

**(i) – (dG/dT) _{P} (ii) (dG/dP)_{T} (iii) – (dA/dV)_{T} (iv) – (dA/dT)_{V}**

(A) (i) and (ii)

(B) (i) and (iv)

(C) (i) only

(D) (ii) only

Thermodynamics Q.9 |

** Answer:** (B) (i) and (iv)

**Hints:** we know that a system subjected to only P-V work... **dG = - TdS** at constant Pressure P,

**dS = - (dG/dT)p**

and **dA = - TdS** at constant Volume V,

**dS = - (dA/dT)v**

follow those condition in (i) and (iv)

**Q.10. On reacting 1.55 g of a diol with an excess of methyl magnesium iodide, 1.12 L (corrected to STP) of methane gas is liberated. The molecular mass (g mol-1) of the diol is ____________.**

Thermodynamics Q.10 |

** Answer:** 62 g.mol-1

**Hints:**

**Q.11. Of the following inequalities, the criterion/criteria for spontaneity of a chemical reaction is/are (i) (Î”G) _{T,P} < 0 (ii) (Î”U)_{S,V} > 0 (iii) (Î”S)_{U,V} > 0**

(A) (i) only

(B)(ii) only

(C) (i) and (ii)

(D) (i) and (iii)

Thermodynamics Q.11 |

**Answer:** (D) (i) and (iii)

**Hints:** At a spontaneity of a chemical reaction must follow those conditions (i) **(Î”G) _{T, P} < 0**

(ii)** (Î”U) _{S,V} < 0**

(iii) **(Î”S) _{U,V} > 0**

**Q.12. For an elementary bimolecular gas phase reaction, activation energy is 5.5 kJ mol-1. Enthalpy of activation, in kJ mol-1, at 300 K is _____________. (R = 8.314 J K-1 mol-1)**

Thermodynamics Q.12 |

**Answer:** 0.512 KJ

**Hints:**

For an **elementary bimolecular gas-phase reaction**, ** Activation energy (Ea)** and

**relation is**

*Enthalpy activation (Î”H#)***E_a=〖Î”H〗^#+2RT**

5.5×1000=〖Î”H〗^#+2×8.314×300

〖Î”H〗^#=5500-4988=512 J.〖mol〗^(-1)=**0.512 KJ.〖mol〗^(-1)**

**Q.13. One mole of an ideal gas is compressed from 5 L to 2 L at a constant temperature. The change in entropy, in J K-1, of the gas is ____________. (R = 8.314 J K-1 mol-1)**

Thermodynamics Q.13 |

**Answer:** -7.618 J.K-1

**Hints:** At constant temperature, Î”U = 0,

According to thermodynamics first low we get, **Î”U = Q + W** or 0 = Q + W or **W = - Q**.

Î”S=Q/T

Î”S=-W/T

**Î”S= nRln V_2/V_1**

Î”S= 1 mol ×8.314×ln 2/5

**Î”S=-7.61 J/K**

**Q.14. The dependence of rate constant k on temperature T (in K) of a reaction is given by the expression**

**lnk=[(-5000K)/T]+10**

**The activation energy of the reaction (in KJ mol-1) is ____________.**

Thermodynamics Q.14 |

**Answer:** 41.57 KJ

**Hints:**

**lnk=[(-5000K)/T]+10**………… (1)

Both sides differenced equation (1) with respect to T, we get

dlnk/dT=((-5000K))/T^2 …………(2)

According to Arrhenius equation …

k=〖Ae〗^(- E_a/RT)…………… (3)

Where A= Arrhenius constant and k =rate constant, Ea =Activation energy at T (in K) temperature.

Both sides take to log in this equation (3), we get

**lnk=lnA-E_a/RT** ………….. (4)

Both sides differenced equation (4) with respect to T, we get

dlnk/dT=0-E_a/(RT^2 )

dlnk/dT=-E_a/(RT^2 )……………. (5)

From eqution (2) and (5) we get,

-E_a/(RT^2 )=((-5000K))/T^2

**E_a=5000×R**

Activation energy E_a=5000×8.314=**41570 J**=41.57 KJ

**Q.15. Consider an ideal gas of volume V at temperature T and pressure P. If the entropy of the gas is S, the partial derivative (dP/dS) _{V} is equal to**

(A) (dT/dP)_{S} (B) (dT/dV)_{P} (C) – (dT/dV)_{S} (D) (dT/dS)_{P}

Thermodynamics Q.15 |

**Answer:** (C) – (dT/dV)S

**Hints:** we know that **(dP/dS)V = – (dT/dV)S**

This equation easy to prove it

According to **Maxwell Relation**, we know that **dU = TdS – PdV** ………. (1)

If S is constant then dS = 0, equation (1) differenced concerning V, at S constant. We get

〖(dU/dV)〗_S=-P…………. (2)

Again, differenced equation (3) concerning S, we get

d/dS dU/dV=-〖(dP/dS)〗_V ………….. (3)

If S is constant then dV = 0, equation (1) differenced concerning S, at V constant. We get

〖(dU/dS)〗_V=T …………… (4)

Again, differenced equation (4) with respect to V, we get

d/dV dU/dS=〖(dT/dV)〗_S ………….. (5)

Now, from equation (3) and (5) we get,

**d/dS dU/dV=d/dV dU/dS**

-(dP/dS)_V=(dT/dV)_S

**Prove it.**

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