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11.Thermodynamics
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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 temperahture is nearly..... $K$ $(R\, = 8.3\, J\,K^{-1}\, mol^{-1} )$
A
$40$
B
$33$
C
$20$
D
$14$
(JEE MAIN-2014)
Solution
$Given:Work\,done,W=830J$
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$
Standard 11
Physics
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