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Two gases have the same initial pressure, volume and temperatue. They expand to the same final volume, one adiabatically and the other isothermally
The final temperature is greater for the isothermal process
The final pressure is greater for the isothermal process
The work done by the gas is greater for the isothermal process
All of the above
Solution
Work done is equal to area under the P-V graph, thus work done by gas is greater for isothermal process.
For isothermal process: $T_{2}=T_{o}$
For adibatic process: $T_{o}\left(V_{o}\right)^{\gamma-1}=T(V)^{\gamma-1}$
As $V>V_{o} \Longrightarrow T<T_{o}$
Thus final temp. is greater for isothermal process.
For isothermal process: $P_{o} V_{o}=P_{i} V \quad \Longrightarrow P_{i}=P_{o} \frac{V_{o}}{V}$
For adibatic process: $P_{o}\left(V_{o}\right)^{\gamma}=P_{a}(V)^{\gamma}$
$\Longrightarrow P_{a}=P_{o}\left(\frac{V_{o}}{V}\right)^{\gamma}$
As $\gamma>1 \quad$ (always) $\Longrightarrow P_{i}>P_{a}$
