A gas absorbs 100 J of heat and is simultaneously compressed by a constant external pressure of 1.50 atmospheres from a volume of 8 L to 2 L. E will be

 

One mole of an ideal gas at 300Kis expanded isothermally from an initial volume of 1 litre to 10 litres. (R = 2 cal mol^–1 K^–1). E for the process is

 

Calculate the work performed when 2 moles of hydrogen expand isothermally and reversibly at 25ºC from 15 to 50 litres.

 

A gas absorbs 200J of heat and expands against the external pressure of 1.5 atm from a volume of 0.5 litre to 1.0 litre, Calculate the change in internal energy-

 

The internal energy of one mole of a gas is

 

A system does 100J work on surroundings by absorbing 150J of heat. Calculate the change in internal energy-

 

A sample of gas is compressed by an average pressure of 0.50 atmosphere so as to decrease its volume from 400 cm^3 to 200 cm3^.During the process 8.00 J of heat flows out to surroundings. Calculate the change in internal energy of the system.

 

A sample of an ideal gas undergoes a process (4 atm, 40 L) to (1 atm, 150 L). If the internal energy of gas decreases by 10 L-atm, H for the gas is

 

5 moles of an ideal gas (Cp,m= 2/7 R) is heated from 27ºC to 87ºC. The value of H is.

 

90 gm water is completely converted into stream at 100ºC and 1 atm. The internal energy of system (latent heat of vaporisation of water at 100ºC= 540 cal/gm)

 

Two moles of an ideal monoatomic gas is heated from 27ºC to 127ºC.The change internal energy of gas is:

 

500 gm ice at 0ºC is added in 2000 gm water at tºC. If the final temperature of system is 0ºC, then the value of 't' is (latent heat of fusion of ice =80 cal/gm and specific heat capacity of water= 1 cal/gm-ºC)

 

2 moles of an ideal gas absorbs 720 cal heat when heated from 27ºC to 87ºC, at constant volume. 'Cp/Cv' for the gas is :

 

Five moles of a monoatomic ideal gas is heated from 300 K to 400 K at constant pressure. The amount of heat absorbed is :

 

Calculate the amount of heat required to raise the temperature of 50 gm water from 25ºC to 55ºC. Specific heat capacity of water = 4.2 J/ºC-gm.

 

One mole of argon gas (assuming ideal behaviour) is expanded reversibly and adiabatically from 2.5 m^3 at 300K temperature to 5.0 m^3. Cv m for argon is 12.55 JK^–1mol^–1.Calculate the final temperature and pressure of the gas.

 

Two mol of an ideal gas is expanded isothermally and reversibly at 300K to twice the original volume. Calculate q, w, E, and H.

 

Heat, Heat Exchange and Heat capacity (C) :

Heat is defined as the energy that flow into or out of the system because of a difference in temperature between system and surrounding.

Note; Work is more organized way of energy transfer as compared to hear exchange.

IUPAC  Sign convention of Heat: 

Sign of heat will negative (-Ve) if heat is released by the system given by system.

While sign of heat will be positive (+Ve) if heat is given to the system.

(i) q_V = n C_v dT (for constant volume process)

(ii) q_p = n C_p dT (for constant pressure process)

(iii) (C_m)p – (C_m)v = R

(iv) C_v & C_p depends on temperature even for an ideal gas.( C = a + bT + cT² .....)

(v) It is a path function

(vi) C_v, C_p are heat capacity of system and (C_m)v  & (C_m)p are heat capacity of one mole system at constant  volume and pressure respectively.

“Exchange of heat and work(P–V) in between system and surrounding always

   occur through boundary of system”

  Note: heat exchange can be measured with the help of Heat Capacity.

 HEAT CAPACITY(C)

 We know that usually on increasing in temperature is proportional to the heat transfer

                               q = coefficient x ∆T

The coefficient depend on the size, composition and nature of the system so we can also write it as   

                                 q= C∆T

 where C is called Heat capacity

The heat capacity (C):
                                     q=C∆T  or

                                   C= q/∆T      unit- J/K

                            If ΔT=1⁰=1K

                                Then C=q

It is equal to amount of heat needed to raise the temperature of the sample of any substance by one degree Celsius (or Kelvin).

heat Capacity depend  on quantity, nature as well as physical state of the system. And the heat capacity is extensive. it may be made intensive as specific heat capacity

                                  

Specific Heat Capacity(C_s ): 

                                     q=C_sm∆T  or

                                   C_s= q/m∆T      

                            If ΔT=1⁰=1K and m = 1g

                                Then C_s=q

It is equal to amount of heat required to raise the temperature of 1gm substance by one degree centigrade .it is intensive properties.

Molar Heat Capacity(C_m ):
                                     q=C_mn∆T  or

                                   C_m= q/n∆T      

                            If ΔT=1⁰=1K and n=1mole

                                Then C_m=q

It is equal to amount of heat required to raise the temperature of 1 Mole substance by one degree centigrade

EXAMPLE(1): The latent heat of fusion of ice at 0ºCis 80 cal/gm the amount of heat needed to convert 200 gm ice into water at 0ºC is ?

 (A) 80 cal (B) 16000 J (C) 16000 cal (D) 1600 cal

SOLUTION: Ans (C)  q = m.L = 200 × 80 = 16000 cal

EXAMPLE(2): Calculate the amount of heat required to raise the temperature of 50 gm water from 25ºC to 55ºC.Specific heat capacity of water = 4.2 J/ºC-gm.

(A) 126 J    (B) 210 J    (C) 6300 J   (D) 1500 J

SOLUTION: Ans. (C)

  q = m.s.ΔT = 50 × 4.2 × (55 – 25) = 6300 J

EXAMPLE (3): Five moles of a monatomic ideal gas is heated from 300K to 400K at constant pressure. the amount of heat absorbed is :

(A) 500 cal   (B) 1500 cal   (C) 2500 cal   (D) 2500 J

SOLUTION: Ans. (C)     

q_p = C_pΔT = n.C_p,m ΔT

    = 5 × 5/2× (400 – 300) = 2500 cal

EXAMPLE (4):  2 moles of an ideal gas absorbs 720 cal heat when heated from 27ºC to 87ºC, at constant volume. 'ɤ' for the gas is :

(A) 1.5 (B) 1.4 (C) 1.6 (D) 1.33

SOLUTION :Ans. (D)

 qv = n.Cv,m. ΔT

C_v,m =qv/nΔT

       = 720/2x(87 – 27)

       = 6 cal/K-mol

Now, r = 1 +R/C_v,m

           = 1 + 2/6

           = 1.33

EXAMPLE (5): 500 gm ice at 0ºC is added in 2000 gm water at tºC. If the final temperature of system is 0ºC, then the value of 't' is (latent heat of fusion of ice = 80 cal/gm and specific heat capacity of water=cal/gm-ºC)

(A)   20    (B)  40      (C) 10     (D)     2

SOLUTION: Ans. (A)

Heat lost by water = heat gained by ice

or, (m.s. ΔT)water = (m.L)ice

or, 2000 × 1 × (t – 0) = 500 × 800

Δ t = 20ºC

EXAMPLE(6): What is the heat in Joules required to raise the temperature of 25 grams of water from 0 °C to 100 °C? What is the heat in calories?

(Given: specific heat of water = 4.18 J/g·°C)

SOLUTION: Use the formula   q = mcΔT
where
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature

q = 25gx4.18 J/g·°x(100 °C - 0 °C)
q = 25gx4.18 J/g·°Cx(100 °C)
q = 10450 J

 We know 1 Calorie=4.18 J

So  10450 J in Calorie   = 10450/4.8=2500 calorie 

RELATION BETWEEN Cp AND Cv :

FOR LARGE HEAT CHANGE :

Case –(1)  when C_m is constant

              Q= nC_m (T₂-T₁)

Case- (2) C_m = f(T)

                 Cm = a + bT+ cT^{2 +……..}

Case- (3) The theoretical value of C_vm and C_pm  for Ideal gas can determined by using degree of freedom.

CHARACTERISTIC OF HEAT CAPACITY:

(1): The heat capacity Of any system should depend upon temperature because by increasing    temperature  of system different degree of freedom get excited.

(2): When temperature approaches zero then heat absorbed by the solid mainly converted into vibration potential energy of molecule resulting in very small increase in temperature, hence ‘C’ increases sharply with increase in temperature.

                   Normally C Directly proportional T³

 (3): When the temperature at Melting point of solid ,then heat capacity becomes nearly  constant  for solid elements.

                    Molar heat capacity= 6.4 Cal/K mole

              Or specific heat capacity x atomic weight=6.4 (Dulong and petite’s law)

(4): Exactly at melting point, the heat capacity become infinite as  ΔT=0

(5): the heat capacity of liquid is greater than that of solid because of rotational degree of freedom also excited.

(6): In liquid heat capacity also depend upon temperature and also infinite at boiling point.

(7): the heat capacity of gas become less than liquid because all vibrational and rotational degree of freedom converted into translational degree of freedom.

(8): the  heat capacity of gases depend upon their atomicity.

(9): If the heat capacity depends upon temperature.

(10); As heat (q) is path function, any substance may have infinite heat capacity.

         Example for any substance.

         Isothermal process = infinite

         Adiabatic process   = 0

         Isobaric process     = C_p

         Isochoric process   = C_v

      Normally ,we use C_p and _Cv  value  as characteristic of substance.

Degree of Freedom and Law of Equipartion of Energy:

It is equal to number of modes of energy transfer when a gaseous molecule undergoes collision.                      OR

        It represent the number of independent modes to describe the molecular motion.  

       Total degree of freedom =  3N (Where N  is Number of atom in molecule)

1- Translational degree of freedom is 3 (three) always for mono, di and t     tri atomic molecule.

2- Rotational degree of freedom is zero for mono atomic gas, 2 (two) for diatomic molecules and   3 (three) for triatomic molecule

3-Vibrational degree of freedom is also zero for mono atomic gas and 1(one) diatomic gas molecule  and for polyatomic gases Vibrational DOF is calculated individually.( f_vib= 3N- f_trans+ f_rot)

Total degree of freedom:=  f_trans + f _rot + f _{vib    &}  f_vib= 3N- f_trans + f _rot    

Molecules                N                      TDF

He                             1                         3

O₂                                    2                         6

CO₂                           3                         9

NH₃                            4                        12

PCl₅                           6                        18

Case-1

Monoatomic       Diatomic       Triatomic (linear)       Triatomic (Non linear)

f _total =3                f_total  =6          f_total =9                      f _total=9

f _trans=3                  f_trans =3            f_trans=3                     f_trans=3

f _rot  =0                      f_rot     =2           f_rot    = 2                    f_rot    =3

f _vib  =0                     f_vib     =1           f_vib    = 4                        f_vib   =3

                              q =n C_mdT

                                q_V=n C_vmdT

                                (C_m)v=(dq/dT)_v

 By FLOT dq= dU+ dW     

At constant volume  dW=0   so   dqv=dU

                         Hence   ( C_m)V= (dU/dT)_v

LAW OF EQUIPARTIAL OF ENERGY :
Average energy associate with each molecule per degree of freedom is U= 1/2KT  (where K is Boltz’s man constant.

Let degree of freedom is = f   then U is U=1/2fkT

             And U=1/2fkTN_A per molecule  we know  kN_A=R

                     U=1/2fRT  and   dU/dT=1/2fR

             And  dU/dT=C_v      hence  C_v=1/2fR

Cv=1/2f_transR +1/2f_rotR  (Where Vib degree inactive in chemistry)

       For ideal gas C_pm-C_vm=R  and  Gama= C_pm/C_vm

         

Adiabatic exponent :Adiabatic exponent (Gama) for a mixture of gas with different heat capacity is defined as :

where n1, n2 ........................ are moles of different gases

Example: Calculate change in internal energy of 10 gm of H₂ ,when it's state is changed from(300K, 1Atm) to (500 K, 2Atm) ?

Solution: For ideal gas

       

C_v for H₂ (diatomic) in low temperature range will be 5R as vibrational part is not included.

Extensive properties

The parameter  whose value change on division known as extensive properties and these are depends on the mass (size, quantity) of the system.

1. Volume  
2. Number of moles     
3. Mass
4. Mole 
5. Free Energy (G)
6. Entropy (S)      
7. Enthalpy (H)   
8. Internal energy (E&U)   
9. Heat capacity
10. K.E.
11. P.E.
12. Gibbs free energy (G)
13. Resistance
14. Conductance

Extensive and Intensive properties:

1: Extensive properties are additive but intensive properties are non additive.

2: Ratio of two extensive property gives an intensive property.

3: An extensive property can be converted into intensive property by defining it per mole/per gram/ per liter