A litre of dry air at STP expands adiabatically to a volume of 3 L. If gamma = 1.40, the work done b

Q: Why does the pressure drop so much faster than in an isothermal process?

In an isothermal process, heat is added to keep the temperature constant as the gas expands. In an adiabatic process, no heat is added. The gas must use its own internal energy to do the work of expanding, which causes its temperature to drop. This drop in temperature, combined with the increase in volume, causes a steeper drop in pressure.

Q: Why can't we just use $W = P\Delta V$?

The formula $W = P\Delta V$ only works for isobaric processes, where the pressure remains perfectly constant throughout the expansion. In this adiabatic process, the pressure is continuously changing, so we must use the integrated formula $W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}$.

Q: What exactly is $\gamma$ (gamma)?

Gamma ($\gamma$) is the adiabatic index, defined as the ratio of the specific heat at constant pressure ($C_p$) to the specific heat at constant volume ($C_v$). For diatomic gases like the nitrogen and oxygen that make up air, $\gamma$ is approximately 1.4.

Q: Why is my calculated answer slightly different from the exact option?

Small differences usually arise from the exact conversion factors used by the examiner. For instance, using $1 \text{ atm} = 1.01 \times 10^5 \text{ Pa}$ versus $101325 \text{ Pa}$, or rounding intermediate steps like $P_2$ to two decimal places. Always choose the closest option.

Q: Is the temperature constant in this process?

No, absolutely not. Because the gas does work on its surroundings without any heat input ($Q = 0$), the First Law of Thermodynamics ($\Delta U = Q - W$) dictates that its internal energy must decrease. For an ideal gas, a decrease in internal energy means a decrease in temperature.

Question

A litre of dry air at STP expands adiabatically to a volume of 3 L. If $\gamma = 1.40$, the work done by air is ($3^{1.4} = 4.6555$)
[Take, air to be an ideal gas]

Options:

(A) 100.8 J

(B) 90.5 J

(C) 48 J

(D) 60.7 J

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Expansion

Imagine you have a perfectly insulated cylinder containing exactly one litre of dry air at Standard Temperature and Pressure (STP). The insulation is crucial here—it means that as the gas expands, absolutely no heat can enter or leave the system. This is the defining characteristic of an **adiabatic process**.

As the gas pushes against the piston to expand its volume from $1 	ext{ L}$ to $3 	ext{ L}$, it must do work. But where does the energy for this work come from if no heat is added? It comes directly from the gas's own internal energy. Consequently, the temperature of the gas drops, and the pressure plummets much faster than it would in a simple isothermal (constant temperature) expansion.

### The Master Equation: Adiabatic Process

To calculate the work done, we first need to know the final state of the gas. Specifically, we need the final pressure, $P_2$. For an adiabatic process involving an ideal gas, the relationship between pressure and volume is governed by Poisson's equation:

$$P_1 V_1^\gamma = P_2 V_2^\gamma$$

Here, $\gamma$ (gamma) is the ratio of specific heats ($C_p/C_v$), which is given as $1.40$ for air (a diatomic gas). We know our initial conditions from STP: $P_1 = 1 	ext{ atm}$ and $V_1 = 1 	ext{ L}$. Our final volume is $V_2 = 3 	ext{ L}$. Let's substitute these values into our master equation:

$$1 	imes (1)^{1.4} = P_2 	imes (3)^{1.4}$$

### Calculating the Final Pressure

Solving for $P_2$ requires evaluating $3^{1.4}$. The problem kindly provides this value: $3^{1.4} = 4.6555$.

$$P_2 = \frac{1}{4.6555} \approx 0.2148 	ext{ atm}$$

Notice how drastically the pressure has dropped! If this were an isothermal process, the pressure would simply be $1/3 \approx 0.33 	ext{ atm}$. The adiabatic pressure drop is much steeper because the gas is also cooling down.

### Calculating the Work Done

The work done by the gas during an adiabatic expansion is the area under the $P-V$ curve. Mathematically, integrating $P \, dV$ using our adiabatic relation yields a beautiful, closed-form formula:

$$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$$

Let's plug in our pressures (in atm) and volumes (in L):

$$W = \frac{(1 	imes 1) - (0.2148 	imes 3)}{1.40 - 1}$$

$$W = \frac{1 - 0.6444}{0.4} = \frac{0.3556}{0.4} = 0.889 	ext{ atm}\cdot	ext{L}$$

### The Final Conversion

We have the work done, but it's in units of atmosphere-litres. The options are in Joules. To convert, we use the conversion factor $1 	ext{ atm}\cdot	ext{L} \approx 101.325 	ext{ J}$.

$$W = 0.889 	imes 101.325 \approx 90.07 	ext{ J}$$

Looking at our options, $90.5 	ext{ J}$ is the closest match. The slight discrepancy arises from the exact approximations used by the examiner (often taking $1 	ext{ atm} = 1.01 	imes 10^5 	ext{ Pa}$, which makes $1 	ext{ atm}\cdot	ext{L} = 101 	ext{ J}$, and rounding $P_2$ slightly differently). Regardless, the physics is rock solid, and the correct choice is undeniably clear.

Animated Solution for Physics - Thermodynamics: A litre of dry air at STP expands adiabatically to a volume of 3 L. If $γ = 1.40$ , the work done by air is ( $3^{1.4} = 4.6555$ ) [Take, air to be an ideal gas]

Select Answer:

Visualized Solution

Visualizing the Expansion

Adiabatic Process Equation

Substituting Known Values

Calculating $P_{2}$

Work Done Formula

Substituting Values for Work

Evaluating Work in atm-L

Converting to Joules

The Sigma Insight: Thermodynamic Processes

The Setup

The Master Equation

Calculating the Final Pressure

Calculating the Work Done

The Final Conversion

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