Thermodynamics chemistry solved problem

Solved Problems on Thermodynamics:-

Problem 1:-

A container holds a mixture of three nonreacting gases: n1 moles of the first gas with molar specific heat at constant volume C1, 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 ΔT.

C = Q/ΔT

So, Q = C ΔT

Each species will experience the equal temperature change.

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

Q = nC ΔT

Solution:-

If the gas has n1 moles, then the amount of heat energy Q1 transferred to a body having heat capacity C1 will be,

Q1 = n1C1 ΔT

Similarly, if the gas has n2 moles, then the amount of heat energy Q2 transferred to a body having heat capacity C2 will be,

Q2 = n2C2 ΔT

And

if the gas has n3 moles, then the amount of heat energy Q3 transferred to a body having heat capacity C3 will be,

Q3 = n3C3 ΔT

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

Q = Q1 + Q2 + Q3

   = n1C1 ΔT + n2C2 ΔT + n3C3 ΔT

Dividing both the sides by n = n1 + n2 + n3 and ΔT, then we will get,

Q/nΔT = (n1C1 ΔT + n2C2 ΔT + n3C3 ΔT)/ nΔT

As, Q/nΔT = C, thus,

          C=ΔT (n1C1 + n2C2  + n3C3)/ nΔT

          = (n1C1 + n2C2  + n3C3)/ n

          = n1C1 + n2C2  + n3C3/ n1 + n2 + n3

From the above observation we conclude that, the molar specific heat at constant volume of the mixture would be n1C1 + n2C2  + n3C3/ n1 + n2 + n3.

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 = ΔEint

Here Q is the energy transferred between the system and its environment, W is the work done on or by the system and ΔEint 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 ΔT.

C = Q/ΔT

So, Q = C ΔT

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/mΔT

So, Q = c mΔT

Solution:-

The heat transfers for the water Qw is,

Qw = mwcw (Tf –Ti)

Here, mass of water is mw, specific heat capacity of water is cw, final temperature is Tf and initial temperature is Ti.

The heat transfers for the thermometer Qt is

Qt = CtΔTt

Here, heat capacity of thermometer is Ct and ΔTt is the temperature difference.

As the internal energy of the system is zero and there is no work is done, therefore substitute ΔEint = 0 and W = 0 in the equation Q + W = ΔEint,

Q + W = ΔEint

 Q + 0= 0

So, Q = 0

Or, Qw + Qt = 0

mwcw (Tf –Ti)+ CtΔTt = 0

So, Ti = (mwcw Tf + CtΔTt )/ mwcw

Here ΔTt = 44.4 ° C - 15.0 ° C

               =  29.4 ° C

To obtain the temperature of the water before insertion Ti of the thermometer, substitute 0.3 kg for mw, 4190 J/kg.m for cw, 44.4 ° C for Tf, 46.1 J/K for Ct and 29.4 ° C for ΔTt in the equation Ti = (mwcw Tf + CtΔTt )/ mwcw,

Ti = (mwcw Tf + CtΔTt )/ mwcw

   = [(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 ΔS for a reversible isothermal process is defined as,

ΔS = 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 mw  and mass of final state ms.

So, m = mw - ms

To obtain mass of water that changed into ice m, substitute 1.78 kg for mass of water mw and 1.02 kg for mass of final state ms in the equation m = mw - ms,

m = mw - ms

   = 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 ΔS of the system during this process, substitute 0.76 kg for mass m, 333×103 J/kg for heat of fusion of water L and 273 K for T in the equation ΔS = -mL/T,

ΔS = -mL/T

     = -(0.76 kg) (333×103 J/kg )/(273 K)

     = -927 J/K

From the above observation we conclude that, the change in entropy ΔS 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 ΔS of the system during this process is equal to the negative of previous case.

So, ΔS = -(- 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 ΔS is always      zero.

The total change in entropy will be,

ΔS = (-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 = TL / TH - TL        …… (1)

Here TL is the lower temperature of sink and TH is the higher temperature of source.


A refrigerator would like to extract as much heat QL 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)

   = QL/W

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

Thus, W = QL/K      …… (2)

Substitute the value of K from equation (1) in the equation W = QL/K,

W = QL/K

    = QL/( TL / TH - TL)

    = QL (TH/ TL – 1)       …… (3)


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


W = QH – QL

Here QH is the exhausted heat.

Thus exhausted heat will be,

QH = W + QL     …… (4)

Substitute the value of W from equation (3) in the equation QH = W + QL,

QH = W + QL    

      = QL (TH/ TL – 1) + QL

     = QL (TH/ TL)       


Solution:-


To obtain the minimum amount of heat delivered to the laboratory, substitute 150 mJ for QL, 296 K for TH and 4.0 K for TL in the equation QH = QL (TH/ TL),

QH = QL (TH/ TL)

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

      = 11 

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