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+1 vote

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 : $\;(R=8.3JK^{-1}mol^{-1})\;$

$(a)\;40\;K\qquad(b)\;30\;K\qquad(c)\;20\;K\qquad(d)\;14\;K$

Can you answer this question?
 
 

1 Answer

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Best answer

The work done in adiabatic process is given as
$$W=\dfrac{nR\Delta T}{\gamma-1}$$
or
$$830=\dfrac{2\times 8.3\times \Delta T}{1.4-1}$$
or
$$\Delta T=20  K$$

answered Mar 3, 2015 by ashu.gaikwad.56
selected Mar 3, 2015 by pady_1
 

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