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  • Can someone help me with this, - Thermodynamics - BITSAT

For a reversible process at T=300k, the volume of 2 mole of ideal gas is increased from 1 litre to 10 litres, the \delta H for isothermal change is:

  • Option 1)

    11.47 kJ

  • Option 2)

    4.98 kJ

  • Option 3)

    0

  • Option 4)

    2.49 kJ

 

Answers (1)

best_answer

As learnt in

First law of Thermodynamics -

Energy of universe is always conserved or total energy of an isolated system is always conserved

\Delta E= q + W


 

- wherein

\Delta E= Internal Energy

q= Heat

W= work

 

 and

Reversible Isobaric Process -

W=P\left ( V_{f}-V_{i} \right )
 

- wherein

P_{gas}= constant

Since pressure of gas is defined during whole of the process and hence the process must be reversible

 

 For an isothermal process, \Delta E =0

\Delta H =\Delta E+P\Delta V

\Delta H =P\Delta V = nR\Delta T\ \ \ \ \ \left ( \Delta T=0 \right )

\therefore \Delta H = 0


Option 1)

11.47 kJ

Option 2)

4.98 kJ

Option 3)

0

Option 4)

2.49 kJ

Posted by

Plabita

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