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Heat and Thermodynamics question

2015 · Q154
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Heat and Thermodynamics question

2015 · Q154

NEETPhysicsHeat and ThermodynamicsMCQ+4 / −1
One mole of an ideal diatomic gas undergoes a transition from A to B along a path AB as shown in the figure.

AIPMT 2015 Cancelled Paper Physics - Heat and Thermodynamics Question 69 English

The change in internal energy of the gas during the transition is
  1. A
    20 J
  2. B
    −-− 12 kJ
  3. C
    20 kJ
  4. D
    −-− 20 kJ
View written solutionFree

Correct answer: D

Change in internal energy from A →\to→ B

ΔU=f2nRΔT=f2nR(Tf−Ti)\Delta U = {f \over 2}nR\Delta T = {f \over 2}nR\left( {{T_f} - {T_i}} \right)ΔU=2f​nRΔT=2f​nR(Tf​−Ti​)

=52{PfVf−PiVi} = {5 \over 2}\left\{ {{P_f}{V_f} - {P_i}{V_i}} \right\}=25​{Pf​Vf​−Pi​Vi​}

(As gas is diatomic ∴\therefore∴ f = 5)

=52{2×103×6−5×103×4} = {5 \over 2}\left\{ {2 \times {{10}^3} \times 6 - 5 \times {{10}^3} \times 4} \right\}=25​{2×103×6−5×103×4}

=52{12−20}×103J=5×(−4)×103J = {5 \over 2}\left\{ {12 - 20} \right\} \times {10^3}J = 5 \times \left( { - 4} \right) \times {10^3}J=25​{12−20}×103J=5×(−4)×103J

∴ΔU=−20 KJ \therefore \Delta U = - 20\,KJ∴ΔU=−20KJ

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