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Chapter 2: Problem 11

A cylinder of compressed gas has a cross-sectional area of \(50 \mathrm{~cm}^{2}\). How much work is done by the system as the gas expands, moving the piston \(15 \mathrm{~cm}\) against an external pressure of \(121 \mathrm{kPa}\) ?

Short Answer

Expert verified
The work done is 90.75 Joules.

Step by step solution

01

Understand the Problem

The problem involves finding the work done by a gas as it expands and moves a piston. We are given the cross-sectional area of the cylinder, the distance the piston moves, and the external pressure.
02

Calculate Volume Change

The formula for the volume change \(\Delta V\) in a cylinder is given by \(\Delta V = A \times d\), where \(A\) is the cross-sectional area and \(d\) is the distance the piston travels. Here, \(A = 50\, \mathrm{cm}^2\) and \(d = 15\, \mathrm{cm}\). Thus, \[ \Delta V = 50 \times 15 = 750 \, \mathrm{cm}^3. \]
03

Convert Units of Volume

Since pressure is given in kPa and requires m鲁 for compatibility, convert \(\Delta V\) from \(\mathrm{cm}^3\) to \(\mathrm{m}^3\). \(1 \, \mathrm{m}^3 = 10^6 \, \mathrm{cm}^3\), so \[ \Delta V = \frac{750}{10^6} = 0.00075 \, \mathrm{m}^3. \]
04

Apply the Formula for Work Done

Work done by the gas during expansion can be calculated using the formula \(W = P_{\text{ext}} \times \Delta V\), where \(P_{\text{ext}} = 121 \, \mathrm{kPa}\) or \(121,000 \, \mathrm{Pa}\). Substitute the values into the formula: \[ W = 121,000 \times 0.00075 = 90.75 \, \mathrm{J}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Work Done by Gas
When gas expands in a cylinder, it performs work. This work is essentially the energy used by the gas to push the piston against an opposing force鈥攁lso known as the external pressure. The work done by the gas can be calculated using the fundamental equation: \[ W = P_{\text{ext}} \times \Delta V, \]where:
  • \( W \) is the work done by the gas,
  • \( P_{\text{ext}} \) is the external pressure, and
  • \( \Delta V \) is the change in volume.
Essentially, the work done by the gas is the product of the constant pressure and the volume change. This formula helps us quantify the energy transfer in thermodynamic processes.
External Pressure
External pressure is the resisting force exerted by something outside the system鈥攊n this context, acting against the gas trying to expand. External pressure is critical because:
  • It determines how much work has to be done by the gas.
  • This pressure can come from atmospheric pressure or any other application pushing back on the piston.
In the exercise, the external pressure was given as \(121 \ \mathrm{kPa}\). Before using it in calculations, it needs conversion into pascals (Pa), since most work calculations in physics utilize the SI unit of pascal. The conversion is straightforward: \[121 \, \mathrm{kPa} = 121,000 \, \mathrm{Pa}.\]The accurate conversion ensures the work calculation will be compatible with standard units.
Volume Change
When we discuss volume change in a cylinder with gas, we are essentially measuring how much the gas expands or contracts. The volume change \( \Delta V \) can be calculated by using the formula:\[ \Delta V = A \times d, \]where:
  • \( A \) is the cross-sectional area,
  • \( d \) is the distance the piston travels.
In the provided problem, with \( A = 50 \, \mathrm{cm}^2 \) and \( d = 15 \, \mathrm{cm} \), the volume change is computed as \( 750 \, \mathrm{cm}^3 \). This value is crucial for determining the amount of work done as the gas expands against an external pressure.
Unit Conversion
Unit conversion is an essential step in solving physics problems accurately, especially in thermodynamics. Often, given values are not in the units needed for the final calculation. For instance, volume in cubic centimeters \( \mathrm{cm}^3 \) must be converted to cubic meters \( \mathrm{m}^3 \) since standard thermodynamic equations require SI units. This is how you do it for volume:
  • Remember, \( 1 \mathrm{m}^3 = 10^6 \mathrm{cm}^3 \).
  • To convert, divide the volume in \( \mathrm{cm}^3 \) by \( 10^6 \).
  • In our problem, \( 750 \, \mathrm{cm}^3 \) becomes \( 0.00075 \, \mathrm{m}^3 \).
Accurate unit conversion ensures all quantities fit correctly into the equations, leading to meaningful results.

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