# 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

P Plabita

As we learnt in

Total internal energy -

$U= U_{K}+U_{P}$

- wherein

Change in internal energy

$\Delta U= n\frac{f}{2}R\Delta T$

(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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