Question
One mole of an ideal diatomic gas expands from volume V to 2 V isothermally at a temperature $27^{\circ}C$ and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from $27^{\circ}C$ doing the same amount of work W , then its final temperature will be (close to) $\_\_\_\_$ $\ ^{\circ}C$.
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Solution
For Isothermal process
$$\begin{matrix} W_{\text{isothermal~}} & & \ = nRT\mathcal{l}n\left( \frac{{\text{ }V}_{2}}{{\text{ }V}_{1}} \right) \\ = & & 1.R300.\mathcal{l}n(2) \\ = & 300R(0.693)\ldots. & \text{(1)} \end{matrix}$$
Now for adiabatic process,
It is given work done in Isothermal = work done in adiabatic
$$\begin{array}{r} W_{\text{adiabatic~}} = \frac{nR\left( T_{1} - T_{2} \right)}{\gamma - 1}\#(2) \end{array}$$
$${(1) = (2) }{\frac{nR\left( 300 - T_{\text{final~}} \right)}{1.4 - 1} = 300R(0.693) }{T_{\text{Final~}} = 216.84\text{ }K }{= - {56.3}^{\circ}C}$$
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