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Step-by-Step Solution
Step 1: Identify the Type of Process
The gas undergoes an isothermal compression (temperature remains constant) from 5 m3 to 1 m3 against a constant external pressure of 4 N·m–2.
Step 2: Calculate the Work Done
The work done in an irreversible isothermal process against constant external pressure is given by:
$ W = -P_{\text{ext}} (V_{2} - V_{1}) $
$P_{\text{ext}} = 4$ N·m–2
Initial volume, $V_{1} = 5$ m3
Final volume, $V_{2} = 1$ m3
Substitute the values:
$ W = -4 \times (1 - 5) = -4 \times (-4) = 16 \text{ J} $
The positive value means 16 J of work is done on the gas.
Step 3: Relate Heat and Work for an Isothermal Ideal Gas Process
For an isothermal process involving an ideal gas, the change in internal energy, $ \Delta U $, is zero:
$ \Delta U = 0
\quad\Rightarrow\quad
q = -W $
Because $W = 16$ J is done on the gas, the heat released by the gas is $ q = -16$ J. This means 16 J of heat is given off to the surroundings (in this case, used to heat the aluminum).
Step 4: Use the Heat Released to Raise the Temperature of Aluminum
The heat released is absorbed by 1 mole of Al, raising its temperature. If $ n = 1 $ mole, molar heat capacity $ C_m = 24 $ J·mol–1·K–1, and the heat gained by Al is $ q = 16 $ J, then:
$ q = n \times C_m \times \Delta T
= 1 \times 24 \times \Delta T
$
So,
$
16 = 24 \times \Delta T
\quad\Rightarrow\quad
\Delta T = \frac{16}{24} = \frac{2}{3} \text{ K}.
$
Step 5: State the Final Answer
The temperature of Al increases by
$ \displaystyle \frac{2}{3} \text{ K}.
$
