Temperature of 1 mole of an ideal gas is increased from 300K to 310K under isochoric process. Heat supplied to the gas in this process is Q = 25 R, where R = universal gas constant. The amount of work to be done by the gas, if temperature of the gas decreases from 310K to 300K adiabatically, is ...............R.

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answer is 25.

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Detailed Solution

$\begin{array}{l} {ΔQ}_{v} = {nC}_{v} ΔT \\ \Rightarrow 25 R = (1) (C_{v}) (310 - 300) \\ \Rightarrow C_{v} = \frac{5}{2} R \\ \Rightarrow γ = 1 .4 \end{array}$
Now work done in adiabatic process,

$W = \frac{nR (T_{1} - T_{2})}{γ - 1} = \frac{(1) (R) (310 - 300)}{1 .4 - 1} = 25 R$

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