**Solved Problems on Thermodynamics:-**

**Problem 1:-**

A container holds a mixture of three nonreacting gases: n_{1} moles of the first gas with molar specific heat at constant volume C_{1}, and so on. Find the molar specific heat at constant volume of the mixture, in terms of the molar specific heats and quantitites of the three separate gases.

**Concept:-**

Heat capacity *C* of a body as the ratio of the amount of heat energy *Q* transferred to a body in any process to its corresponding temperature change Δ*T*.

*C* = *Q*/Δ*T*

So, *Q* = *C* Δ*T*

Each species will experience the equal temperature change.

If the gas has *n* molecules, then *Q* will be,

*Q* = *nC* Δ*T*

**Solution:-**

If the gas has *n*_{1} moles, then the amount of heat energy *Q*_{1} transferred to a body having heat capacity* C*_{1} will be,

*Q*_{1} = *n*_{1}*C*_{1} Δ*T*

Similarly, if the gas has *n*_{2} moles, then the amount of heat energy *Q*_{2} transferred to a body having heat capacity* C*_{2} will be,

*Q*_{2} = *n*_{2}*C*_{2} Δ*T*

And

if the gas has *n*_{3} moles, then the amount of heat energy *Q*_{3} transferred to a body having heat capacity* C*_{3} will be,

*Q*_{3} = *n*_{3}*C*_{3} Δ*T*

As, each species will experience the same temperature change, thus,

*Q* = *Q*_{1} + *Q*_{2} + *Q*_{3}

= *n*_{1}*C*_{1} Δ*T* + *n*_{2}*C*_{2} Δ*T* +* n*_{3}*C*_{3} Δ*T*

Dividing both the sides by *n* = *n*_{1} + *n*_{2} + *n*_{3} and Δ*T*, then we will get,

*Q*/*n*Δ*T* = (*n*_{1}*C*_{1} Δ*T* + *n*_{2}*C*_{2} Δ*T* +* n*_{3}*C*_{3} Δ*T*)/* n*Δ*T*

As,* Q*/*n*Δ*T* = *C*, thus,

*C*=Δ*T* (*n*_{1}*C*_{1} + *n*_{2}*C*_{2} +* n*_{3}*C*_{3})/* n*Δ*T*

= (*n*_{1}*C*_{1} + *n*_{2}*C*_{2} +* n*_{3}*C*_{3})/* n*

= *n*_{1}*C*_{1} + *n*_{2}*C*_{2} +* n*_{3}*C*_{3}/* n*_{1} +* n*_{2} +* n*_{3}

From the above observation we conclude that, the molar specific heat at constant volume of the mixture would be *n*_{1}*C*_{1} + *n*_{2}*C*_{2} +* n*_{3}*C*_{3}/* n*_{1} +* n*_{2} +* n*_{3.}

**Problem 2:-**

A thermometer of mass 0.055 kg and heat capacity 46.1 J/K reads 15.0°C. It is then completely immersed in 0.300 kg of water and it comes to the same final temperature as the water. If the thermometer reads 44.4°C, what was the temperature of the water berfore insertion of the thermometer, neglecting other heat losses?

**Concept:-**

In accordance to the law of conservation of energy, for a thermodynamic system, in which internal is the only type of energy the system may have, the law of conservation of energy may be expressed as,

*Q* + *W* = Δ*E*_{int}

Here Q is the energy transferred between the system and its environment, W is the work done on or by the system and Δ*E*_{int} is the change in the internal energy of the system.

The heat capacity *C* of a body as the ratio of amount of heat energy *Q* transferred to a body in any process to its corresponding temperature change Δ*T*.

*C* = *Q*/Δ*T*

So, *Q* = *C* Δ*T*

The heat capacity per unit mass of a body, called specific heat capacity or usually just specific heat, is characteristic of the material of which the body is composed.

*c *= *C*/*m*

= *Q*/*m*Δ*T*

So, *Q* = *c m*Δ*T*

**Solution:-**

The heat transfers for the water* Q*_{w} is,

*Q*_{w} = *m*_{w}*c*_{w} (*T*_{f} –*T*_{i})

Here, mass of water is *m*_{w}, specific heat capacity of water is *c*_{w}, final temperature is *T*_{f} and initial temperature is *T*_{i}.

The heat transfers for the thermometer* Q*_{t} is

*Q*_{t} = *C*_{t}Δ*T*_{t}

Here, heat capacity of thermometer is *C*_{t} and Δ*T*_{t} is the temperature difference.

As the internal energy of the system is zero and there is no work is done, therefore substitute Δ*E*_{int} = 0 and *W* = 0 in the equation *Q* + *W* = Δ*E*_{int},

*Q* + *W* = Δ*E*_{int}

*Q* + 0= 0

So, *Q* = 0

Or, *Q*_{w} + *Q*_{t} = 0

*m*_{w}*c*_{w} (*T*_{f} –*T*_{i})+* C*_{t}Δ*T*_{t} = 0

So, *T*_{i} = (*m*_{w}*c*_{w} *T*_{f} + *C*_{t}Δ*T*_{t} )/* m*_{w}*c*_{w}

Here Δ*T*_{t} = 44.4 ^{°} C - 15.0 ^{°} C

= 29.4 ^{°} C

To obtain the temperature of the water before insertion* T*_{i} of the thermometer, substitute 0.3 kg for *m*_{w}, 4190 J/kg.m for *c*_{w}, 44.4 ^{°} C for *T*_{f}, 46.1 J/K for *C*_{t} and 29.4 ^{°} C for Δ*T*_{t} in the equation *T*_{i} = (*m*_{w}*c*_{w} *T*_{f} + *C*_{t}Δ*T*_{t} )/* m*_{w}*c*_{w,}

*T*_{i} = (*m*_{w}*c*_{w} *T*_{f} + *C*_{t}Δ*T*_{t} )/* m*_{w}*c*_{w}

= [(0.3 kg) (4190 J/kg.m) (44.4 ^{°} C) + (46.1 J/K) (29.4 ^{°} C)] /[(0.3 kg) (4190 J/kg.m)]

=45.5 ^{°} C

From the above observation we conclude that, the temperature of the water before insertion of the thermometer was 45.5 ^{°} C.

**Problem 3:-**

A mixture of 1.78 kg of water and 262 g of ice at 0°C is, in a reversible process, brought to a final equilibrium state where the water / ice ratio, by mass 1:1 at 0°C. (a) Calculate the entropy change of the system during this process. (b) The system is then returned to the first equilibrium state, but in an irreversible way (by using a Bunsen burner, for instance). Calculate the entropy change of the system during this process. (c) Show that your answer is consistent with the second law of thermodynamics.

**Concept:-**

The entropy change Δ*S* for a reversible isothermal process is defined as,

Δ*S* = *Q*/*T*

= -*mL*/*T*

Here *m* is the mass, *L* is the latent heat and *T* is the temperature.

**Solution:-**

(a) Mass of water = 1.78 kg

Mass of ice = 262 g

So the total mass of ice and water mixture will be,

Mass of ice-water mixture = (Mass of water) + (Mass of ice)

= (1.78 kg) + (262 g)

= (1.78 kg) + (262 g×10^{-3} kg/1 g)

= 1.78 kg + 0.262 kg

= 2.04 kg

If eventually the ice and water have the same mass, then the final state will have 1.02 kg (2.04 kg/2) of each.

Thus the mass of the water that changed into ice *m* will be the difference of mass of water *m*_{w } and mass of final state *m*_{s}.

So, *m* = *m*_{w} - *m*_{s}

To obtain mass of water that changed into ice *m*, substitute 1.78 kg for mass of water* m*_{w} and 1.02 kg for mass of final state* m*_{s} in the equation *m* = *m*_{w} - *m*_{s},

*m* = *m*_{w} - *m*_{s}

= 1.78 kg – 1.02 kg

= 0.76 kg

The change of water at 0^{°} C to ice at 0^{°} C is isothermal.

To obtain the change in entropy Δ*S* of the system during this process, substitute 0.76 kg for mass *m*, 333×10^{3} J/kg for heat of fusion of water *L* and 273 K for *T* in the equation Δ*S* = -*mL*/*T*,

Δ*S* = -*mL*/*T*

= -(0.76 kg) (333×10^{3} J/kg )/(273 K)

= -927 J/K

From the above observation we conclude that, the change in entropy Δ*S* of the system during this process will be -927 J/K.

(b) Now the system is returned to the first equilibrium state, but in an irreversible way. Thus the change in entropy Δ*S* of the system during this process is equal to the negative of previous case.

So, Δ*S* = -(- 927 J/K)

= 927 J/K

From the above observation we conclude that, the change in entropy ΔS of the system would be 927 J/K.

(c) In accordance to second law of thermodynamics, entropy change Δ*S* is always zero.

The total change in entropy will be,

Δ*S* = (-927 J/K) + (927 J/K)

= 0

From the above observation we conclude that, our answer is consistent with the second law of thermodynamics.

**Problem 4:-**

Apparatus that liquefies helium is in a laboratory at 296 K. The helium in the apparatus is at 4.0 K. If 150 mJ of heat is transferred from the helium, find the minimum amount of heat delivered to the laboratory.

**Concept:-**

Coefficient of performance *K* of a Carnot refrigerator is defined as,

*K* = *T*_{L} / *T*_{H} - *T*_{L} …… (1)

Here *T*_{L} is the lower temperature of sink and *T*_{H} is the higher temperature of source.

A refrigerator would like to extract as much heat *Q*_{L} as possible from the low-temperature reservoir (“what you want”) for the least amount of work *W* (“what you pay for”). So the efficiency of a refrigerator is defined as,

*K* = (what you want)/(what you pay for)

= *Q*_{L}/*W*

and this is called coefficient of performance. The larger the value of *K*, the more efficient is the refrigerator.

Thus, *W* = *Q*_{L}/*K* …… (2)

Substitute the value of *K* from equation (1) in the equation *W* = *Q*_{L}/*K*,

*W* = *Q*_{L}/*K*

= *Q*_{L}/(* T*_{L} / *T*_{H} - *T*_{L})

= *Q*_{L} (*T*_{H}/* T*_{L} – 1) …… (3)

The first law of thermodynamics, applied to the working substance of the refrigerator, gives,

*W* = *Q*_{H} – *Q*_{L}

Here *Q*_{H} is the exhausted heat.

Thus exhausted heat will be,

*Q*_{H} = *W* + *Q*_{L} …… (4)

Substitute the value of *W* from equation (3) in the equation *Q*_{H} = *W* + *Q*_{L},

*Q*_{H} = *W* + *Q*_{L}

= *Q*_{L} (*T*_{H}/* T*_{L} – 1) + *Q*_{L}

= *Q*_{L} (*T*_{H}/* T*_{L})

**Solution:-**

To obtain the minimum amount of heat delivered to the laboratory, substitute 150 mJ for *Q*_{L}, 296 K for *T*_{H} and 4.0 K for *T*_{L} in the equation *Q*_{H} = *Q*_{L} (*T*_{H}/* T*_{L}),

*Q*_{H} = *Q*_{L} (*T*_{H}/* T*_{L})

= ((150 mJ) (10^{-3} J/1 mJ)) (296 K/4.0 K)

= 11 J

From the above observation we conclude that, the minimum amount of heat delivered to the laboratory would be 11 J.

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