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The work of 146 kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7^{\circ}C . The gas is (R=8.3\, J\, mol^{-1}K^{-1})

  • Option 1)

    monoatomic

  • Option 2)

    diatomic

  • Option 3)

    triatomic

  • Option 4)

    a mixture of monoatomic and diatomic.

 

Answers (1)

best_answer

As we learnt in 

Adiabatic Process -

When a Thermodynamic System undergoes a change in such a way that no exchange of heat takes place.

- wherein

\Delta Q= 0

 

and

 

Work done in Adiabatic process -

W = \int PdV
 

- wherein

W = \frac{nR\left ( T_{i}-T_{f} \right )}{\gamma -1}

\gamma = adiabatic\, \, exponent

 

For adiabatic process

W = Change in internal energy

= n C_{v}\left ( T_{i}-T_{f} \right )

146\times 10^{3} = \left ( 1000 \right ) \times \frac{R}{\gamma - 1} \times 7

146 = \frac{8.3\times 7}{\gamma -1}

or \gamma -1 = \frac{58.1}{146}

\gamma = 1.4

\therefore gas is diatomic  

 

 

 


Option 1)

monoatomic

This option is incorrect.

Option 2)

diatomic

This option is correct.

Option 3)

triatomic

This option is incorrect.

Option 4)

a mixture of monoatomic and diatomic.

This option is incorrect.

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Plabita

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