5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature

Q: Why can we use $\Delta U = nC_V\Delta T$ even though the volume is changing during compression?

For an ideal gas, internal energy is a state function that depends *only* on temperature, not on volume or pressure. Therefore, the formula $\Delta U = nC_V\Delta T$ holds true for *any* process involving an ideal gas, whether the volume is constant or not.

Q: Why did we use $R = 8.0$ instead of the standard $8.314$?

In competitive exams, examiners sometimes provide simplified values for constants to make calculations faster and test your conceptual understanding rather than your arithmetic stamina. Always use the exact values given in the problem statement.

Q: Does the fact that the compression is 'reversible' change our calculations?

No. Both internal energy $U$ and the product $pV$ are state functions. This means their changes depend only on the initial and final states (temperatures), regardless of the path taken (reversible or irreversible) to get there.

Q: Could we have found $\Delta U$ using the First Law of Thermodynamics, $\Delta U = q + w$?

While the First Law is always valid, we don't have enough information here. We don't know the heat exchanged $q$ or the work done $w$ directly. Using the state function property $\Delta U = nC_V\Delta T$ is the direct and intended method.

Q: What if the gas wasn't ideal? Would these formulas still work?

No. For real gases, internal energy depends on both temperature and volume due to intermolecular forces. The relation $\Delta U = nC_V\Delta T$ and the ideal gas law $pV = nRT$ are specific approximations that only apply when gas molecules are assumed to have no volume and no attractive forces.

Question

5 moles of an ideal gas at $100	ext{ K}$ are allowed to undergo reversible compression till its temperature becomes $200	ext{ K}$. If $C_V = 28	ext{ JK}^{-1}	ext{mol}^{-1}$, calculate $\Delta U$ and $\Delta(pV)$ for this process. ($R = 8.0	ext{ JK}^{-1}	ext{mol}^{-1}$)

Options:

(A) $\Delta U = 2.8	ext{ kJ}; \Delta(pV) = 0.8	ext{ kJ}$

(B) $\Delta U = 14	ext{ J}; \Delta(pV) = 0.8	ext{ J}$

(C) $\Delta U = 14	ext{ kJ}; \Delta(pV) = 4	ext{ kJ}$

(D) $\Delta U = 14	ext{ kJ}; \Delta(pV) = 18	ext{ kJ}$

Sigma Percentile · Accepted Answer

### Visualizing the State Change

Imagine a sturdy cylinder fitted with a movable piston. Inside this cylinder, we have trapped exactly $5 	ext{ moles}$ of an ideal gas.

Initially, the gas is resting at a cool temperature of $100 	ext{ K}$.

Now, we slowly and reversibly push the piston down, compressing the gas. As we do work on the gas, its temperature rises, eventually reaching $200 	ext{ K}$. This is our physical setup.

### Calculating Internal Energy Change

The first thing we need to determine is the change in the internal energy of the gas, denoted by $\Delta U$.

For an ideal gas, the internal energy is a direct reflection of its temperature. There are no intermolecular forces to worry about, so the energy is purely kinetic.

Because of this, the change in internal energy depends **only** on the change in temperature, regardless of what happens to the pressure or volume. The master formula we use is:

$$ \Delta U = n C_V \Delta T $$

Here, $n$ is the number of moles, $C_V$ is the molar heat capacity at constant volume, and $\Delta T$ is the change in temperature.

Let's plug in our known values. We have $n = 5 	ext{ mol}$, $C_V = 28 	ext{ JK}^{-1}	ext{mol}^{-1}$, and our temperature change $\Delta T$ is $(200 - 100) 	ext{ K}$.

$$ \Delta U = 5 	imes 28 	imes (200 - 100) $$

$$ \Delta U = 140 	imes 100 = 14000 	ext{ J} $$

Converting this to kilojoules, we get our first crucial result:

$$ \Delta U = 14 	ext{ kJ} $$

### The Ideal Gas Law and pV Change

Next, we need to find the change in the product of pressure and volume, $\Delta(pV)$.

At first glance, this might seem tricky because we don't know the initial or final pressures or volumes. However, we have a powerful tool at our disposal: the **Ideal Gas Equation**.

$$ pV = nRT $$

Since the amount of gas $n$ and the gas constant $R$ are fixed, any change in the $pV$ product must be directly proportional to the change in temperature.

Therefore, we can write:

$$ \Delta(pV) = \Delta(nRT) = nR\Delta T $$

This elegant substitution saves us from needing to know the individual pressures and volumes.

Let's substitute the values provided in the problem. We use $n = 5 	ext{ mol}$, the given $R = 8.0 	ext{ JK}^{-1}	ext{mol}^{-1}$, and $\Delta T = 100 	ext{ K}$.

$$ \Delta(pV) = 5 	imes 8.0 	imes 100 $$

$$ \Delta(pV) = 40 	imes 100 = 4000 	ext{ J} $$

Converting to kilojoules, we find:

$$ \Delta(pV) = 4 	ext{ kJ} $$

### Conclusion

By systematically applying the principles of thermodynamics and the ideal gas law, we have successfully determined both required quantities.

The change in internal energy is **$14 	ext{ kJ}$**, and the change in the $pV$ product is **$4 	ext{ kJ}$**.

This perfectly matches option (c), confirming our analytical approach.

Animated Solution for Chemistry - Chemical Thermodynamics: 5 moles of an ideal gas at $100 K$ are allowed to undergo reversible compression till its temperature becomes $200 K$ . If $C_{V} = 28 JK^{- 1} mol^{- 1}$ , calculate $Δ U$ and $Δ (p V)$ for this process. ( $R = 8.0 JK^{- 1} mol^{- 1}$ )

Select Answer:

Visualized Solution

$Visualizing the Process$

$Formula for Δ U$

$Substituting Values for Δ U$

$Calculating Δ U$

$Formula for Δ (p V)$

$Substituting Values for Δ (p V)$

$Calculating Δ (p V)$

$Final Answer$

The Sigma Insight: First Law of Thermodynamics

Visualizing the State Change

Calculating Internal Energy Change

The Ideal Gas Law and pV Change

Conclusion

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For an ideal gas

Select Answer:

Visualized Solution

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formula for Δ(pV)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for Δ(pV)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Δ(pV)

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