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Molar and Specific Heat Capacity

The work done...

Question

The work done by one mole of ideal gas during an adiabatic process is not given by:

w = C_{v} (T_{f} - T_{i})

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w = \frac{P_{f} V_{f} - P_{i} V_{i}}{γ - 1}

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w = \frac{P_{f} V_{f} - P_{i} V_{i}}{γ - 1} [1 - (\frac{P_{i}}{P_{f}})^{\frac{γ - 1}{γ}}]

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w = \frac{P_{f} V_{f} - P_{i} V_{i}}{γ - 1} [1 + \frac{P_{f}}{P_{i}}]^{γ - 1}

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Solution

The correct option is D $w = \frac{P_{f} V_{f} - P_{i} V_{i}}{γ - 1} [1 + \frac{P_{f}}{P_{i}}]^{γ - 1}$
$w = \frac{n R}{γ - 1} \times [T_{f} - T_{i}]$
$\frac{n R}{(γ - 1)} \times \frac{P_{f} V_{f} - P_{i} V_{i}}{n R} = \frac{P_{f} . V_{f} - P_{i} . V_{i}}{γ - 1} (∵ P V = n R T)$
$\frac{P_{f} . V_{f}}{γ - 1} [1 - [\frac{P_{i} . V_{i}}{P_{f} . V_{f}}]]$
$\frac{P_{f} . V_{f}}{γ - 1} [1 - \frac{P_{i}}{P_{f}} \times [\frac{P_{f}}{P_{i}}]^{\frac{1}{γ}}] [∵ {P V}^{γ} = c o n s t a n t]$
$\frac{P_{f} . V_{f}}{γ - 1} ⎡ ⎢ ⎢ ⎣ 1 - \frac{P_{i}^{1 - \frac{1}{γ}}}{P_{f}^{1 - \frac{1}{γ}}} ⎤ ⎥ ⎥ ⎦$
$\frac{P_{f} . V_{f}}{γ - 1} [1 - [\frac{P_{i}}{P_{f}}]^{\frac{γ - 1}{γ}}]$

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