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Q.

Work done in raising temperature of five mole of Helium adiabatically through $θ^{\circ} C$ is W. Then the work done for the same change of temperature of 5 mole of hydrogen in adiabatic process is

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a

$\frac{3 W}{5}$

b

$\frac{2 W}{3}$

c

$\frac{3 W}{2}$

d

$\frac{5 W}{3}$

answer is D.

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Detailed Solution

$W = \frac{n R (T_{1} - T_{2})}{γ - 1} a n d W^{'} = n R (T_{2} - T_{1})$

$\frac{W^{1}}{W} = - (γ - 1) = \frac{- 2}{3}$

Case 1:

$W = \frac{n R (T_{1} - T_{2})}{γ - 1} = \frac{5 R (θ)}{\frac{5}{3} - 1} = \frac{15 R θ}{2} . . . . . . . . . (1)$

Case2:

$W' = \frac{n R (T_{1} - T_{2})}{γ - 1} = \frac{5 R (θ)}{\frac{7}{5} - 1} = \frac{25 R θ}{2} . . . . . . . . . (2)$

After solving (1) and (2) we get,

$W' = \frac{25}{2} \times \frac{2 W}{15} = \frac{5 W}{3}$

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