Problem 95 - Chemistry

Question

A certain gas initially at \(0.050 \mathrm{~L}\) undergoes expansion until its volume is \(0.50 \mathrm{~L}\). Calculate the work done (in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 0.20 atm. (The conversion factor is \(1 \mathrm{~L} \cdot \mathrm{atm}=101.3 \mathrm{~J} .\) )

Step-by-Step Solution

Answer

(a) 0 J, (b) -9.117 J

Step 1: Understanding the Problem

We're tasked with calculating the work done by a gas when it expands from an initial volume of \(0.050\,\mathrm{L}\) to a final volume of \(0.50\,\mathrm{L}\) under two different conditions: (a) against a vacuum and (b) against a constant pressure of \(0.20\,\mathrm{atm}\).

Step 1: Work Against a Vacuum

When a gas expands against a vacuum, there is no opposing pressure, so the external pressure \( P_\text{ext} = 0\). Consequently, the work done \( W \) is 0 because work is defined as \( W = -P_\text{ext} \times \Delta V\) and any multiplication with zero results in zero. Thus, \( W = 0 \; \mathrm{J}\).

Step 2: Calculate Change in Volume

The change in volume (\(\Delta V\)) is calculated by subtracting the initial volume from the final volume: \( \Delta V = V_f - V_i = 0.50\,\mathrm{L} - 0.050\,\mathrm{L} = 0.45\,\mathrm{L} \).

Step 3: Work Against Constant Pressure

When the gas expands against a constant pressure of \(0.20\,\mathrm{atm}\), we use the formula \( W = -P_\text{ext} \times \Delta V\). Substitute \( P_\text{ext} = 0.20\,\mathrm{atm}\) and \( \Delta V = 0.45\,\mathrm{L}\) to get: \( W = -0.20\,\mathrm{atm} \times 0.45\,\mathrm{L} = -0.090\,\mathrm{L}\cdot \mathrm{atm}\).

Step 4: Convert Work to Joules

Use the conversion factor \(1\,\mathrm{L}\cdot\mathrm{atm} = 101.3\,\mathrm{J}\) to convert the work from \(\mathrm{L}\cdot\mathrm{atm}\) to joules: \( W = -0.090\,\mathrm{L}\cdot\mathrm{atm} \times 101.3\,\mathrm{J/L}\cdot\mathrm{atm} = -9.117\,\mathrm{J}\).

Key Concepts

Gas Expansion

Gas expansion is a crucial concept in thermodynamics. It happens when a gas increases in volume. This can occur for various reasons, such as an increase in temperature or a decrease in pressure. When gas expands, the molecules spread out and occupy more space. Gas expansion can be categorized into different types based on the conditions under which it occurs:

• **Isothermal Expansion:** The temperature remains constant during the expansion process.
• **Adiabatic Expansion:** The process occurs without any heat exchange with the surroundings.
• **Isobaric Expansion:** The pressure remains constant during the process.

Every type has its own applications and significance. In the given problem, the gas expands from an initial volume of 0.050 L to a final volume of 0.50 L. Whether expanding against a vacuum or constant pressure, the expansion allows us to study the work done by the gas under different conditions.

Constant Pressure Work

When a gas expands against a constant pressure, it performs work. To calculate this work, we use the relationship:\[ W = -P_{\text{ext}} \times \Delta V \]Here, \( P_{\text{ext}} \) is the constant external pressure and \( \Delta V \) is the change in volume. The negative sign indicates that work is done by the gas as it expands.In the original exercise, the external pressure is given as 0.20 atm. Therefore, when the gas expands from 0.050 L to 0.50 L, we use the formula to find the work done in L·atm. Converting this to joules using the conversion factor allows us to understand the energy perspective of the expansion.This concept is instrumental in areas where energy conservation and transfer need to be precisely calculated, such as in engines and other mechanical systems.

Volume Change

Volume change describes the difference between the final volume and the initial volume of the gas. This is mathematically represented as:\[ \Delta V = V_f - V_i \]where \( V_f \) is the final volume and \( V_i \) is the initial volume. In the problem, the gas goes from 0.050 L to 0.50 L, resulting in a volume change of 0.45 L.Understanding volume change is vital because it determines the amount of work a gas does as it expands. It ties the physical process to mathematical expressions, allowing us to quantify the work involved accurately.Recognizing how volume change impacts work calculations is essential in predicting gas behavior in various scenarios. Whether that's against different pressure conditions or in controlled lab settings, knowing \( \Delta V \) helps in making informed decisions about system efficiencies and potential energy outputs.