This problem involves two main concepts: work done during isothermal compression of an ideal gas and heat transfer to raise the temperature of a substance. Let's break it down step by step.

Step 1: Work Done on the Gas

The gas undergoes isothermal compression (constant temperature) against a constant external pressure. The work done on the gas (W) is given by:

$W=-P_{\mathrm{ext}}\times ∆V$

Where:

• $P_{\mathrm{ext}}=4N/m^2$ (external pressure)
• $∆V=V_{\mathrm{final}}-V_{\mathrm{initial}}=1m^3-5m^3=-4m^3$

Substituting the values:

$W=-\left(4\right)\times (-4)=16J$

Since work is done on the gas, W is positive (16 J).

Step 2: Heat Released

For an ideal gas undergoing an isothermal process, the change in internal energy (∆U) is zero. From the first law of thermodynamics:

$∆U=Q+W$

Where Q is heat added to the system. Since ∆U = 0:

$0=Q+W\Rightarrow Q=-W=-16J$

Negative Q means heat is released by the gas. So, heat released = 16 J.

Step 3: Temperature Increase of Aluminum

This released heat (16 J) is used to raise the temperature of 1 mole of Al. The heat required to change temperature is given by:

$Q=n\times C\times ∆T$

Where:

• n = number of moles = 1
• C = molar heat capacity = 24 J mol⁻¹ K⁻¹
• ∆T = change in temperature (to find)

So,

$16=1\times 24\times ∆T$

Solving for ∆T:

$∆T=\frac{16}{24}=\frac{2}{3}K$

Final Answer

The temperature of Al increases by $\frac{2}{3}$ K.

Related Topics and Formulae

First Law of Thermodynamics

$∆U=Q+W$

Where ∆U is change in internal energy, Q is heat added to the system, and W is work done on the system.

Work Done at Constant Pressure

For compression/expansion against constant external pressure: $W=-P_{\mathrm{ext}}∆V$

Heat Capacity

Heat required to change temperature: $Q=nC∆T$ (for molar heat capacity C).

Ideal Gas Isothermal Process

For an ideal gas, internal energy depends only on temperature. So, for isothermal process (∆T=0), ∆U=0.