Work done in reversible adiabatic process in given by:

2.303 RT log \frac{V_{2}}{V_{1}}

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\frac{n R}{(γ - 1)} (T_{2} - T_{1})

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2.303 RT log \frac{V_{1}}{V_{2}}

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none of these

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Solution

The correct option is B $\frac{n R}{(γ - 1)} (T_{2} - T_{1})$
According to first law of thermodyamics $d U = d q + d w$ for adiabatic precess dq = 0. Hence $d U = d w$ .....(i)
and $d U = C_{v} d T$ .....(ii)
Also $\frac{C_{p}}{C_{v}} = γ$
$\frac{C_{v} + n R}{C_{v}} = γ ∵ C_{p} - C_{v} = n R$
$C_{v} = \frac{n R}{γ - 1}$ .....(iii)
Using equation (ii) and (iii)
$d U = \frac{n R}{γ - 1} d T$ ....(iv)
Using equation (i) and (iv) we get
Work done in reversible adiabatic process $(w) = Δ U = \frac{n R}{(γ - 1)} (T_{2} - T_{1})$

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Similar questions

Table - A	Table - B
(a) Work done in isobaric process	(d) $n R T {log}_{e} (\frac{V_{2}}{V_{1}})$
(b) Wok done in isothermal process	(e) $P (V_{2} - V_{1})$
(c) Work done in adiabatic process	(f) $\frac{n R T (T_{1} - T_{2})}{γ - 1}$
	(g) Zero

{N H}_{3}

27^{o} C

γ = 1.33

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