Q 1 :    

Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following:            [2024]

  • q=0,ΔT0,w=0

     

  • q=0,ΔT<0,w0

     

  • q0,ΔT=0,w=0

     

  • q=0,ΔT=0,w=0

     

(D)        For adiabatic process q=0.

              W=-PextΔV, for free expansion, Pext=0, so w=0

              By first law of thermodynamics:

               ΔU=W+q

               ΔU=0+0=0

              For an ideal gas internal energy is function of temperature, so if ΔU=0,ΔT is also 0.



Q 2 :    

If 5 moles of an ideal gas expands from 10 L to a volume of 100 L at 300 K under isothermal and reversible condition then work, w, is -x J. The value of x is _________ .

 

(Given R = 8.314 J K-1 mol-1)                 [2024]



(28721)

For isothermal reversible process, work W is given by:         

     W=-2.303 nRT log V2V1

          =-2.303×5 mol×8.314Jmol K×300 K log 10010

          =-28720.713 J-28721J

 



Q 3 :    

The heat of combustion of solid benzoic acid at constant volume is –321.30 kJ at 27°C. The heat of combustion at constant pressure is (–321.30 – xR) kJ, the value of x is _________ .             [2024]



(150)

C6H5COOH(s)+152O2(g)7CO2(g)+3H2O(l)

Δng=Gaseous moles of products-Gaseous moles of reactants

         =7-7.5=-0.5

Heat of combustion at constant pressure (ΔH) is related to heat of combustion at constant volume (ΔU) by:

ΔH=ΔU+ΔngRT

ΔH=-321.3-0.5R×300

ΔH=-321.3-150R

 



Q 4 :    

ΔvapHΘ for water is + 40.79 kJ mol-1 at 1 bar and 100°C. Change in internal energy for this vapourisation under same condition is ________ kJ mol-1. (Integer answer)

 

(Given R = 8.3 JK-1 mol-1)                                                   [2024]



(38)

H2O(l)H2O(g)

Δng=Gaseous moles of products-Gaseous moles of reactants

         =1-0=1

ΔH=ΔU+ΔngRT

40.79kJmol-1=ΔU+1×8.3JmolK×373K

40.79kJmol-1=ΔU+1×8.31000kJmolK×373K

ΔU=37.69kJmol-138kJmol-1

 



Q 5 :    

If three moles of an ideal gas at 300 K expand isothermally from 30 dm3 to 45 dm3 against a constant opposing pressure of 80 kPa, then the amount of heat transferred is _____ J.                   [2024]



(1200)

[IMAGE 8]

Work (W) against constant pressure is calculated as:

W=-Pext(V2-V1)=-80kPa×(45-30)dm3

                                      =-80×1000Pa×15×(10-1m)3

                                      =-1200 J

By first law of thermodynamics:

ΔU=Q+W

For isothermal process of an ideal gas, ΔU=0

So, Q=-W

Q=-(-1200J)=1200J

Hence 1200 J of heat is absorbed in the process.



Q 6 :    

[IMAGE 9]

An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path ABCA as shown in the diagram above. The total work done in the process is _____ J.           [2024]



(200)

Magnitude of work done in a cyclic process = area enclosed by P-V or V-P graph.

|W|= 12×AC×AB 

         =12×(30-10)kPa×(30-10)dm3

         =200 kPadm3

         =200×1000Pa×(10-1m)3

          =200Pam3=200J

 



Q 7 :    

[IMAGE 10]

Consider the figure provided.

1 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18°C. If the piston is moved to position B, keeping the temperature unchanged, then ‘x’ L atm work is done in this reversible process.

x = _____ L atm. (nearest integer)

[Given: Absolute temperature = °C + 273.15,
R = 0.08206 L atm mol-1 K-1]                          [2024]



(55)

Work done by gas in isothermal reversible expansion is:

w=-2.303nRTlog10V2V1

w=-2.303×1×0.08206×291.15×log1010010 atmL

w=-2.303×1×0.08206×291.15×log1010 atmL

w=-55.023 atmL-55 atmL

Work done by system = +55 atmL