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One mole of a monoatomic gas undergoes adiabatic reversible expansion from volume $V_{1}$ to $V_{2}$ lf 298 K and 278 K are the initial and final temperatures of the gas, the work involved in this expansion process is

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$- (20 mol) R$

$- (50 mol) R$

$- (40 mol) R$

$- (30 mol) R$

answer is B.

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Detailed Solution

$w = n C_{V, m} (T_{2} - T_{1}) = (1 mol) (\frac{3}{2}) R (298 K - 278 K) = - (30 mol) R$

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