The change in the magnitude of the volume of an ideal gas when a small additional pressure Δp is app

Q: Why do we use differentiation for these changes?

The problem specifies "small additional pressure" and "small quantity $\Delta T$". In physics, very small changes ($\Delta$) can be treated mathematically as differentials ($d$), allowing us to apply the product rule to equations like $pV = nRT$.

Q: Why do we take the absolute value or magnitude?

The problem explicitly states "change in the magnitude of the volume". In an isothermal compression, pressure increases ($\Delta p > 0$) while volume decreases ($\Delta V < 0$). Taking magnitudes ensures we are comparing the absolute sizes of these changes without worrying about negative signs.

Q: Can I solve this without calculus?

Yes, you can use finite differences. For isothermal: $p_1V_1 = p_2V_2 \Rightarrow pV = (p+\Delta p)(V-\Delta V)$. Expanding and ignoring the very small $\Delta p \Delta V$ term gives $p\Delta V - V\Delta p = 0$, which is exactly what differentiation gives!

Q: Why did we substitute $\frac{V}{nR}$ with $\frac{T}{p}$?

We were given the numerical values for the initial temperature ($300\text{ K}$) and initial pressure ($2\text{ atm}$), but we didn't know the volume $V$ or the number of moles $n$. Using the ideal gas equation to swap variables allowed us to use the given data directly.

Q: Does the type of gas (monoatomic, diatomic) matter here?

Not at all! The ideal gas equation $pV = nRT$ applies universally to all ideal gases, regardless of their degrees of freedom. The specific heat capacities ($C_p, C_v$) are not involved in this particular process calculation.

Question

The change in the magnitude of the volume of an ideal gas when a small additional pressure $\Delta p$ is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity $\Delta T$ at constant pressure. The initial temperature and pressure of the gas are 300 K and 2 atm, respectively. If $|\Delta T| = C|\Delta p|$, then value of $C$ (in K/atm) is ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing an ideal gas trapped in a container. The gas is initially at a specific state, which we can call point A on a $p-V$ diagram, with a temperature of $300	ext{ K}$ and a pressure of $2	ext{ atm}$.

The problem describes two distinct thought experiments starting from this exact same state. In the first scenario, we gently squeeze the gas, increasing its pressure by a tiny amount $\Delta p$, while keeping the temperature perfectly constant. This is an **isothermal process**. Because the pressure increases, the volume must decrease by some small amount $\Delta V$.

In the second scenario, we start over from point A. This time, we keep the pressure constant but cool the gas down slightly by a temperature $\Delta T$. This is an **isobaric process**. Cooling the gas causes it to contract, so the volume decreases again.

The beautiful catch in this problem is that the **magnitude of the volume change** is identical in both scenarios!

### The Master Equation

To solve this, we need our trusty tool: the ideal gas equation, $pV = nRT$. Since we are dealing with very small changes, we can use the power of calculus and differentiate this equation.

Let's look at the isothermal process first. Since the temperature $T$ is constant, the right side of the equation is a constant. Differentiating gives us:
$p\Delta V + V\Delta p = 0$

We can rearrange this to find the magnitude of the pressure change:
$V\Delta p = -p\Delta V$
$|\Delta p| = p\frac{|\Delta V|}{V}$

Now, let's analyze the isobaric process. Here, the pressure $p$ is constant. Differentiating the ideal gas equation gives:
$p\Delta V = nR\Delta T$

From this, we can easily isolate the magnitude of the temperature change:
$|\Delta T| = \frac{p|\Delta V|}{nR}$

### Final Calculation

We have expressions for both $|\Delta p|$ and $|\Delta T|$, and we know that $|\Delta V|$ is the same for both. Let's divide the two expressions to see what happens:
$\frac{|\Delta T|}{|\Delta p|} = \frac{\frac{p|\Delta V|}{nR}}{p\frac{|\Delta V|}{V}}$

Notice how the $p|\Delta V|$ terms cancel out perfectly! We are left with a very elegant relation:
$\frac{|\Delta T|}{|\Delta p|} = \frac{V}{nR}$

But wait, we can simplify this even further. From the ideal gas equation, we know that $\frac{V}{nR} = \frac{T}{p}$. Substituting this back into our ratio gives:
$|\Delta T| = \frac{T}{p}|\Delta p|$

Now, it's just a matter of plugging in the initial values given in the problem. The initial temperature $T$ is $300	ext{ K}$, and the initial pressure $p$ is $2	ext{ atm}$.
$|\Delta T| = \frac{300}{2}|\Delta p|$
$|\Delta T| = 150|\Delta p|$

The problem states that $|\Delta T| = C|\Delta p|$. By comparing our result with this equation, we can confidently conclude that the constant $C$ is **$150$**.

Enter Numerical Value:

Visualized Solution

$p - V$ Diagram Setup

Isothermal Process

Isobaric Process

Volume Constraint

Isothermal Differentiation

Pressure Change Magnitude

Isobaric Differentiation

Ratio of Changes

Simplifying the Ratio

Substituting Values

Final Answer

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

An ideal gas is expanding such that $p T^{2} = constant$ . The coefficient of volume expansion of the gas is

An ideal gas is expanding such that $p T^{3} = constant$ . The coefficient of volume expansion of the gas is

During an experiment, an ideal gas is found to obey an additional law $p^{2} V = constant$ . The gas is initially at a temperature $T$ and volume $V$ . When it expands to a volume $2 V$ , the temperature becomes ……… .

An ideal gas is initially at temperature $T$ and volume $V$ . Its volume is increased by $Δ V$ due to an increase in temperature $Δ T$ , pressure remaining constant. The quantity $δ = Δ V / (V Δ T)$ varies with temperature as

In a certain thermodynamical process, the pressure of a gas depends on its volume as $k V^{3}$ . The work done when the temperature changes from $10 0^{\circ} C$ to $30 0^{\circ} C$ will be ......... $n R$ , where $n$ denotes number of moles of a gas.

A given quantity of an ideal gas is at pressure $p$ and absolute temperature $T$ . The isothermal bulk modulus of the gas is

During an adiabatic process, the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio $C_{p} / C_{V}$ for the gas is

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation $V T = k$ , where $k$ is a constant. In this process, the temperature of the gas is increased by $Δ T$ . The amount of heat absorbed by gas is (where, $R$ is gas constant)

A diatomic ideal gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of the gas is $T_{i}$ (in kelvin) and the final temperature is $a T_{i}$ , the value of $a$ is