An ideal gas undergoes isothermal compression from 5 m^3 to 1 m^3 against a constant external pressu

Q: Why is $\Delta U = 0$ for the gas?

For an ideal gas, the internal energy depends only on its temperature. Since the process is isothermal, the temperature remains constant, meaning the internal energy does not change.

Q: Why do we use $W = -p_{\text{ext}}\Delta V$?

This is the formula for irreversible work done against a constant external pressure. The negative sign ensures that compression (where $\Delta V$ is negative) results in positive work done *on* the gas.

Q: Why is the heat $q$ for the gas negative?

A negative $q$ indicates that the system (the gas) is losing or releasing heat to its surroundings. This makes sense because compressing the gas does work on it, and to keep the temperature constant, it must shed that energy as heat.

Q: Can we use $C_p$ or $C_v$ for the Aluminum block?

For solids like Aluminum, the volume change upon heating is negligible, so $C_p$ and $C_v$ are practically identical. We just use the given molar heat capacity $C$.

Q: What if the compression was reversible?

If it were reversible, the external pressure would not be constant, and we would calculate work using $W = -nRT \ln(V_f/V_i)$. However, the problem explicitly states a constant external pressure, making it an irreversible process.

Question

An ideal gas undergoes isothermal compression from $5 	ext{ m}^3$ to $1 	ext{ m}^3$ against a constant external pressure of $4 	ext{ Nm}^{-2}$. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is $24 	ext{ J mol}^{-1}	ext{K}^{-1}$, the temperature of Al increases by

Options:

(A) $\frac{3}{2} 	ext{ K}$

(B) $1 	ext{ K}$

(C) $2 	ext{ K}$

(D) $\frac{2}{3} 	ext{ K}$

Sigma Percentile · Accepted Answer

The problem presents a fascinating interplay between mechanical work and thermal energy. We have an ideal gas undergoing isothermal compression, and the heat it releases is entirely absorbed by an Aluminum block. Our goal is to find out how much the temperature of the Aluminum block increases.

### Analyzing the Gas Compression

Let's start by focusing on the ideal gas. The problem states that the gas undergoes an **isothermal compression**. The word "isothermal" is our biggest clue here. It means that the temperature of the gas remains perfectly constant throughout the process.

For an ideal gas, the internal energy depends solely on its temperature. Since the temperature doesn't change, the change in internal energy must be zero:
$$ \Delta U = 0 $$

According to the **First Law of Thermodynamics**, the change in internal energy is the sum of the heat added to the system and the work done on the system:
$$ \Delta U = q + W $$

Substituting our finding, we get:
$$ 0 = q + W \implies q = -W $$

This equation tells us a beautiful physical truth: any work done *on* the gas during compression is entirely released as heat *by* the gas to maintain its constant temperature.

### Calculating the Heat Released

Next, we need to calculate the work done on the gas. The gas is compressed against a **constant external pressure**, which means the process is irreversible. The formula for work done in such a case is:
$$ W = -p_{	ext{ext}} \Delta V = -p_{	ext{ext}} (V_f - V_i) $$

We are given the following values:
- External pressure, $p_{	ext{ext}} = 4 	ext{ Nm}^{-2}$
- Initial volume, $V_i = 5 	ext{ m}^3$
- Final volume, $V_f = 1 	ext{ m}^3$

Let's substitute these into our heat equation:
$$ q = -W = p_{	ext{ext}} (V_f - V_i) $$
$$ q = 4 	imes (1 - 5) $$
$$ q = 4 	imes (-4) = -16 	ext{ J} $$

The negative sign indicates that $16 	ext{ J}$ of heat is **released** by the gas into its surroundings.

### Heating the Aluminum Block

Now, where does this released heat go? The problem states that it is used to increase the temperature of an Aluminum block. Therefore, the heat absorbed by the Aluminum block is exactly the magnitude of the heat released by the gas:
$$ q_{	ext{Al}} = 16 	ext{ J} $$

To find the temperature increase, we use the calorimetry formula:
$$ q_{	ext{Al}} = n \cdot C \cdot \Delta T $$

We are given:
- Number of moles of Al, $n = 1 	ext{ mol}$
- Molar heat capacity of Al, $C = 24 	ext{ J mol}^{-1}	ext{K}^{-1}$

Substituting these values into our equation:
$$ 16 = 1 	imes 24 	imes \Delta T $$

### Final Calculation

All that's left is a simple algebraic step to solve for the change in temperature, $\Delta T$:
$$ \Delta T = \frac{16}{24} 	ext{ K} $$

By dividing the numerator and the denominator by their greatest common divisor, 8, we simplify the fraction:
$$ \Delta T = \frac{2}{3} 	ext{ K} $$

Thus, the temperature of the Aluminum block increases by **$\frac{2}{3} 	ext{ K}$**. This elegant problem perfectly demonstrates how mechanical work done on one system can be quantified and tracked as it transforms into thermal energy in another!

Select Answer:

Visualized Solution

$System Setup$

$Δ U = 0$

$W = - p_{ext} Δ V$

$q_{gas} = - 16 J$

$q_{Al} = n C Δ T$

$16 = 1 \times 24 \times Δ T$

$Δ T = \frac{2}{3} K$

The Sigma Insight: First Law of Thermodynamics

Analyzing the Gas Compression

Calculating the Heat Released

Heating the Aluminum Block

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔU=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } W=−pext​ΔV

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } qgas​=−16 J

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } qAl​=nCΔT

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 16=1×24×ΔT

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ΔT=32​ K