JEE Main 2014 · 11 Apr · Shift 1 · Physics Q12

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During an adiabatic compression, 830 J of work is done on 2 moles of a diatomic ideal gas to reduce its volume by 50%. The change in its temperature is nearly : ( $R=8.3$ JK $^{-1}$ mol $^{-1}$ )

40 K

33 K

20 K

14 K

Answer and worked solution

Official answer: C

In an adiabatic process the work done on the gas all goes into internal energy:
$830=nC_V\Delta T=2\times\tfrac52\times8.3\times\Delta T\implies\Delta T=20\text{ K}.$

Topics:Thermodynamic processes First law and internal energy

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JEE Main 2014 · 11 Apr · Shift 1, Physics question 12