One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 2 7 ∘ C . If the work done during the process is 3 kJ , the final temperature will be equal to ( C v

Question

One mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 2 7 ∘ C . If the work done during the process is 3 kJ , the final temperature will be equal to ( C v ​ = 20 JK − 1 )

TestHub · Accepted Answer

The first law of thermodynamics for an adiabatic process states $d U = d W$ .

For an ideal gas, $d U = n C_{v} d T$ .

Given $d W = - 3 k J = - 3000 J$ (work done BY the gas is negative).

$n C_{v} (T_{2} - T_{1}) = W$

$1 m o l \times 20 J K^{- 1} \times (T_{2} - (27 + 273) K) = - 3000 J$

$20 (T_{2} - 300) = - 3000$

$T_{2} - 300 = - 150$

$T_{2} = 150 K$

Chemistry - Thermodynamics - I Question with Solution | TestHub

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