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JEE Mains · Physics · STD 11 - Thermodynamics
  • JEE Mains 2014
  • Medium
  • Numerical

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 temperahture is nearly..... \(K\) \((R\, = 8.3\, J\,K^{-1}\, mol^{-1} )\)

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Answer & Solution

Correct Answer

(C) \(20\)

Step-by-step Solution

Detailed explanation

\(Given:Work\,done,W=830J\)

no, of moles of gas, \(\mu =2\)

For diatomic gas \(\gamma  = 1.4\)

Work done during an adiabatic change

\(W = \frac{{\mu R\left( {{T_1} - {T_2}} \right)}}{{\gamma  - 1}}\)

\( \Rightarrow 830 = \frac{{2 \times 8.3\left( {\Delta T} \right)}}{{1.4 - 1}} = \frac{{2 \times 8.3\left( {\Delta T} \right)}}{{0.4}}\)

\( \Rightarrow \Delta T = \frac{{830 \times 0.4}}{{2 \times 8.3}} = 20\,K\)

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