Consider one mole of helium gas enclosed in a container at initial pressure P1 and volume V1. It exp

Q: Why is $\gamma = 5/3$?

Helium is a noble gas, which means it is monoatomic. For monoatomic gases, the degrees of freedom are 3, making $\gamma = 1 + 2/f = 1 + 2/3 = 5/3$.

Q: Why use $P_1V_1$ instead of $nRT$ for isothermal work?

Both are perfectly valid since $P_1V_1 = nRT_1$. However, since the problem doesn't give us the temperature $T$, keeping everything in terms of $P_1V_1$ allows the variables to cancel out beautifully when we take the final ratio.

Q: How to calculate $(1/8)^{5/3}$ without a calculator?

Fractional exponents can be split! First, take the cube root (the denominator of the fraction): $(1/8)^{1/3} = 1/2$. Then, raise that result to the 5th power (the numerator): $(1/2)^5 = 1/32$.

Q: Why is the adiabatic work formula $P_iV_i - P_fV_f$ and not the other way around?

Work done BY the gas is positive during expansion. In an adiabatic expansion, the gas cools down, meaning $P_iV_i > P_fV_f$. Writing it this way ensures the work comes out positive without needing a negative sign in front.

Q: What if the gas was diatomic like Oxygen?

If it were diatomic, $\gamma$ would be $7/5$. This would drastically change the final pressure calculation and the denominator in the adiabatic work formula, leading to a completely different ratio.

Question

Consider one mole of helium gas enclosed in a container at initial pressure $P_1$ and volume $V_1$. It expands isothermally to volume $4V_1$. After this, the gas expands adiabatically and its volume becomes $32V_1$. The work done by the gas during isothermal and adiabatic expansion processes are $W_{	ext{iso}}$ and $W_{	ext{adia}}$, respectively. If the ratio $\frac{W_{	ext{iso}}}{W_{	ext{adia}}} = f \ln 2$, then $f$ is ________.

Sigma Percentile · Accepted Answer

### The Tale of Two Expansions: Isothermal vs Adiabatic

Imagine you are tracking the journey of a single mole of helium gas inside a perfectly sealed cylinder. This gas is about to undergo a two-part thermodynamic adventure. Our goal is to calculate the work done in each phase and find their ratio. Let's break down this beautiful interplay of physics and mathematics.

### Phase 1: The Isothermal Stretch

The gas starts at an initial state with pressure $P_1$ and volume $V_1$. In the first phase, it expands **isothermally** until its volume becomes $4V_1$.

Because the process is isothermal, the temperature remains constant. According to Boyle's Law, $P 	imes V = 	ext{constant}$. Since the volume has increased by a factor of 4, the pressure must drop by a factor of 4 to compensate. Therefore, the new pressure is $P' = \frac{P_1}{4}$.

Now, let's calculate the work done during this isothermal expansion. The formula is:

$$W_{	ext{iso}} = nRT \ln\left(\frac{V_f}{V_i}ight)$$

Since $nRT = P_1V_1$, we can substitute this directly:

$$W_{	ext{iso}} = P_1V_1 \ln\left(\frac{4V_1}{V_1}ight) = P_1V_1 \ln(4)$$

Using the power rule of logarithms, $\ln(4) = \ln(2^2) = 2 \ln 2$. Thus, the isothermal work is:

$$W_{	ext{iso}} = 2 P_1V_1 \ln 2$$

### Phase 2: The Adiabatic Plunge

Next, the gas undergoes an **adiabatic** expansion from its new state $(P', 4V_1)$ until its volume reaches a massive $32V_1$. In an adiabatic process, no heat is exchanged, and the governing equation is $PV^\gamma = 	ext{constant}$.

Since helium is a monoatomic gas, its adiabatic index is $\gamma = \frac{5}{3}$. Let's find the final pressure $P_2$:

$$P' (4V_1)^{\frac{5}{3}} = P_2 (32V_1)^{\frac{5}{3}}$$

Substituting $P' = \frac{P_1}{4}$ and rearranging for $P_2$:

$$P_2 = \frac{P_1}{4} \left(\frac{4V_1}{32V_1}ight)^{\frac{5}{3}} = \frac{P_1}{4} \left(\frac{1}{8}ight)^{\frac{5}{3}}$$

To evaluate $\left(\frac{1}{8}ight)^{\frac{5}{3}}$ without a calculator, take the cube root first (which is $\frac{1}{2}$), and then raise it to the 5th power (which gives $\frac{1}{32}$).

$$P_2 = \frac{P_1}{4} 	imes \frac{1}{32} = \frac{P_1}{128}$$

Now, we calculate the work done during this adiabatic phase using the standard formula:

$$W_{	ext{adia}} = \frac{P_i V_i - P_f V_f}{\gamma - 1}$$

Substitute the initial and final states of this specific phase:

$$W_{	ext{adia}} = \frac{\frac{P_1}{4}(4V_1) - \frac{P_1}{128}(32V_1)}{\frac{5}{3} - 1}$$

$$W_{	ext{adia}} = \frac{P_1V_1 - \frac{P_1V_1}{4}}{\frac{2}{3}} = \frac{\frac{3}{4} P_1V_1}{\frac{2}{3}} = \frac{9}{8} P_1V_1$$

### The Grand Finale: Comparing the Work Done

We have successfully calculated the work done in both phases. The final step is to find their ratio:

$$\frac{W_{	ext{iso}}}{W_{	ext{adia}}} = \frac{2 P_1V_1 \ln 2}{\frac{9}{8} P_1V_1}$$

The $P_1V_1$ terms beautifully cancel out, leaving us with:

$$\frac{W_{	ext{iso}}}{W_{	ext{adia}}} = \frac{16}{9} \ln 2$$

The problem states this ratio is equal to $f \ln 2$. By direct comparison, we find that $f = \frac{16}{9}$, which evaluates to approximately **1.78**. A brilliant problem that tests your stamina across multiple thermodynamic states!

Enter Numerical Value:

Visualized Solution

\text{Thermodynamic Processes}

\text{Isothermal Expansion}

\text{Calculating } W_{\text{iso}}

\text{Adiabatic Expansion}

\text{Calculating } P_2

\text{Work Done in Adiabatic Process}

\text{Calculating } W_{\text{adia}}

\text{Finding the Ratio } f

The Sigma Insight: Thermodynamic Processes

The Tale of Two Expansions

Phase 1

Phase 2

The Grand Finale

Similar Questions

Starting with the same initial conditions, an ideal gas expands from volume $V_{1}$ to $V_{2}$ in three different ways, the work done by the gas is $W_{1}$ if the process is purely isothermal, $W_{2}$ if purely isobaric and $W_{3}$ if purely adiabatic, then

5.6 L of helium gas at STP is adiabatically compressed to 0.7 L. Taking the initial temperature to be $T_{1}$ , the work done in the process is

At $2 7^{\circ} C$ two moles of an ideal monoatomic gas occupy a volume $V$ . The gas expands adiabatically to a volume $2 V$ . Calculate (a) the final temperature of the gas, (b) change in its internal energy, (c) the work done by the gas during this process.

An ideal gas expands isothermally from a volume $V_{1}$ to $V_{2}$ and then compressed to original volume $V_{1}$ adiabatically. Initial pressure is $p_{1}$ and final pressure is $p_{3}$ . The total work done is $W$ . Then,

A sample of gas with $γ = 1.5$ is taken through an adiabatic process in which the volume is compressed from $1200 cm^{3}$ to $300 cm^{3}$ . If the initial pressure is $200 kPa$ . The absolute value of the work done by the gas in the process is ......... J.