A gas is expanded from an initial state to a final state along a path on a \(P\)-\(V\) diagram. The path consists of (i) an isothermal expansion of work \(50\;J\), (ii) an adiabatic expansion and (iii) an isothermal expansion of work \(20\;J\). If the internal energy of gas is changed by \(-30\;J\), then the work done by gas during adiabatic expansion is
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For an isothermal process of an ideal gas, \(\Delta U=0\). For an adiabatic process, \(Q=0\), so by first law \(\Delta U=-W\).
Step 1: Recall the first law of thermodynamics. Energy is conserved through $\Delta U = Q - W$, but here it is cleaner to track internal energy directly. The total internal energy change of the gas over the whole path is given as $\Delta U_{total} = -30\ J$. Step 2: Note internal energy in isothermal steps. For an ideal gas, temperature alone fixes internal energy. In an isothermal process the temperature is constant, so $\Delta U = 0$ for each isothermal expansion. Step 3: Conclude where the internal energy changes. Since both isothermal steps contribute zero, the entire internal energy change of $-30\ J$ happens during the adiabatic step. So \[ \Delta U_{adiabatic} = -30\ J \] Step 4: Apply the first law to the adiabatic step. An adiabatic process has no heat exchange, $Q = 0$, so \[ \Delta U = Q - W = -W \] Step 5: Solve for the adiabatic work. \[ -30 = -W \] \[ W = 30\ J \] Step 6: State the answer. The work done by the gas during the adiabatic expansion is $30\ J$, matching option (3). The $50\ J$ and $20\ J$ isothermal works are not needed because they leave internal energy unchanged. \[ \boxed{30\ J} \]
