One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure,

The change in internal energy of the gas during the transition is:

  • Option 1)

    20 J

  • Option 2)

    -12 kJ

  • Option 3)

    20 kj

  • Option 4)

    -20 kJ

 

Answers (1)
P Plabita

As we learnt in 

Total internal energy -

U= U_{K}+U_{P}
 

- wherein

Change in internal energy

Delta U= nfrac{f}{2}RDelta T

(Always)

f is degree of freedom

 

pv =nRT

T = \frac{PV}{nR}

\Delta v =n C_{v}\Delta T = n \frac{R}{r-1}.\left ( T_{f}-T_{i} \right )

or

\Delta v = \frac{1}{r-1}.\left ( nRT_{f} -nRT_{i} \right )

= \frac{P_{f}v_{f}-P_{i}v_{i}}{r-1}

= \frac{2\times 10^{3}\times 6-5\times 10^{3}\times 4}{\frac{7}{5}-1}

= \frac{8\times 10^{3}}{\left ( \frac{2}{5} \right )}J = 20KJ

 


Option 1)

20 J

Incorrect

Option 2)

-12 kJ

Incorrect

Option 3)

20 kj

Correct

Option 4)

-20 kJ

Incorrect

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