Animated Solution for Physics - Thermodynamics: An ideal gas with pressure $p$, volume $V$ and temperature $T$ is expanded isothermally to a volume $2V$ and a final pressure $p_i$. If the same gas is expanded adiabatically to a volume $2V$, the final pressure is $p_a$. The ratio of the specific heats of the gas is $1.67$. The ratio $p_a/p_i$ is .......

Imagine you are in a laboratory, standing in front of a perfectly sealed cylinder fitted with a frictionless piston. Inside this cylinder is an ideal gas. We are going to perform two distinct thermodynamic experiments on this exact same gas, starting from the exact same initial conditions.

Let's define our starting point. The gas initially has a pressure of $p$, a volume of $V$, and a temperature of $T$. Our goal in both experiments is to expand the gas until its volume doubles, reaching $2V$. However, the path we take to get there will be completely different in each case.

In the first experiment, we will expand the gas isothermally. This means we will do it so slowly, and in such good thermal contact with the surroundings, that the temperature $T$ remains absolutely constant. The final pressure in this case will be $p_i$.

In the second experiment, we will expand the gas adiabatically. This means we will wrap the cylinder in perfect insulation so that absolutely no heat can enter or escape. The final pressure in this case will be $p_a$.

Our ultimate mission is to find the ratio of these two final pressures, $\frac{p_a}{p_i}$. To do this, we need to analyze each process individually.

Let's dive into the first experiment: the isothermal expansion. The defining characteristic of an isothermal process is that the temperature remains constant ($\Delta T=0$).

Because the gas is ideal and the temperature is constant, we can invoke one of the most fundamental principles of thermodynamics: Boyle's Law. Boyle's Law states that for a fixed mass of an ideal gas kept at a fixed temperature, the pressure and volume are inversely proportional. Mathematically, the product of pressure and volume is a constant:

Let's plug in the specific values for our isothermal journey. Our initial state is $(p,V)$ and our final state is $(p_i,2V)$. Substituting these into Boyle's Law gives us:

Now, we simply solve for the final isothermal pressure, $p_i$. By dividing both sides by $2V$, the volume $V$ beautifully cancels out:

This result makes perfect intuitive sense. If you double the volume of a gas while keeping its temperature constant, the molecules have twice as much space to move around in. Consequently, they will hit the walls of the container half as often, resulting in exactly half the initial pressure.

Now, let's reset our cylinder to the initial state $(p,V)$ and perform the second experiment: the adiabatic expansion.

In an adiabatic process, the system is thermally isolated. No heat is exchanged with the surroundings ($Q=0$). Because the gas is expanding, it is doing positive work on the piston. According to the First Law of Thermodynamics, this work must come at the expense of the gas's internal energy. As the internal energy drops, the temperature plummets.

Because the temperature is no longer constant, Boyle's Law does not apply here. Instead, we must use the equation of state for a reversible adiabatic process, often known as Poisson's Law:

Here, $\gamma$ (gamma) is the ratio of specific heats ($\frac{C_p}{C_v}$). The problem graciously provides us with the value of $\gamma =1.67$, which tells us we are likely dealing with a monatomic gas like Helium or Argon.

Let's set up Poisson's equation for our adiabatic journey. Our initial state is $(p,V)$ and our final state is $(p_a,2V)$. Equating the initial and final states gives:

We need to isolate the final adiabatic pressure, $p_a$. Let's divide both sides by $(2V{)}^{\gamma }$:

Using the properties of exponents, we can distribute the power of $\gamma$ in the denominator:

Notice how the $V^{\gamma }$ terms perfectly cancel each other out! This leaves us with a clean expression for $p_a$:

Substituting the given value of $\gamma =1.67$, we get:

We have successfully navigated both thermodynamic paths and found expressions for both final pressures. For the isothermal process: $p_i=\frac{p}{2}$ For the adiabatic process: $p_a=\frac{p}{2^{1.67}}$

The question asks for the ratio of the adiabatic final pressure to the isothermal final pressure, $\frac{p_a}{p_i}$. Let's set up the fraction:

When dividing fractions, we multiply by the reciprocal of the denominator:

The initial pressure $p$ cancels out completely. This is a profound realization: the ratio of the final pressures depends only on the expansion ratio and the nature of the gas ($\gamma$), not on the initial pressure itself!

Using the laws of exponents ($\frac{x^a}{x^b}=x^{a-b}$), we can combine the base 2 terms:

All that's left is to evaluate this numerical value.

Calculating $2^{0.67}$ (which is roughly $2^{2/3}$, or the cube root of 4) gives approximately $1.587$.

Our final answer is 0.628.

This result beautifully confirms what we see on a $p-V$ diagram. Because the adiabatic curve is steeper than the isothermal curve, expanding to the same final volume will always result in a lower final pressure for the adiabatic process. Hence, the ratio $\frac{p_a}{p_i}$ must be less than 1, which our mathematical derivation has flawlessly proven.