Chapter 6: Problem 18
One cubic foot of gas under a pressure of 80 pounds per square inch expands adiabatically to 4 cubic feet according to the law \(p v^{1.4}=c\). Find the work done by the gas.
Short Answer
Expert verified
The work done by the gas is approximately 111.6 pound-feet.
Step by step solution
01
Understanding the Problem
We are given that the gas expands adiabatically according to the equation \( pv^{1.4} = c \), starting at 1 cubic foot and expanding to 4 cubic feet, with an initial pressure of 80 psi. We need to find the work done during this expansion.
02
Determine the Constant c
We start by finding the constant \( c \) using the initial conditions. At the initial state, the volume \( v_1 = 1 \) cubic foot and the pressure \( p_1 = 80 \) psi, so \( c = p_1 v_1^{1.4} = 80 \times 1^{1.4} = 80 \).
03
Apply the Law to the Final State
Using the final volume \( v_2 = 4 \) cubic feet and the relation \( pv^{1.4} = 80 \), we find the final pressure \( p_2 \) with \( p_2 = \frac{80}{v_2^{1.4}} = \frac{80}{4^{1.4}} \). Calculating the exponent gives \( 4^{1.4} \approx 8.3 \), so \( p_2 \approx \frac{80}{8.3} = 9.64 \) psi.
04
Calculate the Work Done
The work done by the gas can be calculated using the integral of pressure with respect to volume, \( W = \int_{v_1}^{v_2} p \, dv \). Substituting for \( p = \frac{80}{v^{1.4}} \), the work done is \( W = \int_{1}^{4} \frac{80}{v^{1.4}} \, dv \).
05
Solve the Integral
We solve the integral \( W = 80 \int_{1}^{4} v^{-1.4} \, dv \). The antiderivative is \( F(v) = \frac{80}{-0.4} v^{-0.4} + C \). Evaluating between 1 and 4 gives:\( W = \frac{80}{-0.4} \left[ v^{-0.4} \right]_{1}^{4} = -200 \left[ 4^{-0.4} - 1^{-0.4} \right] \).
06
Compute the Numerical Value
Substituting the values into the expression for work, we have: \( W = -200 [(4^{-0.4}) - 1] \approx -200 [(0.442) - 1] = -200 (-0.558) \approx 111.6 \). The negative sign indicates work done by the gas, so the work is 111.6 pound-feet.
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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 a gas expands, it does work against external forces, such as atmospheric pressure. This work is often quantified in thermodynamic processes, particularly in adiabatic expansions. In this context, the work done by the gas is found by calculating the area under the curve on a pressure-volume (pV) diagram. This represents the integral of the pressure over the change in volume.
For an adiabatic process, where no heat is exchanged with the surroundings, the work done is dependent entirely on the change in internal energy expressed through pressure and volume. Using the formula for work, we can express it as an integral:
\[ W = \int_{v_1}^{v_2} p \, dv \] Where \( v_1 \) and \( v_2 \) are the initial and final volumes, and \( p \) is pressure.
In our adiabatic example, with a fixed \( c = p v^{1.4} \), the pressure changes inversely with volume, which is captured in the integral calculation. The result tells us the total work performed during the expansion process.
For an adiabatic process, where no heat is exchanged with the surroundings, the work done is dependent entirely on the change in internal energy expressed through pressure and volume. Using the formula for work, we can express it as an integral:
\[ W = \int_{v_1}^{v_2} p \, dv \] Where \( v_1 \) and \( v_2 \) are the initial and final volumes, and \( p \) is pressure.
In our adiabatic example, with a fixed \( c = p v^{1.4} \), the pressure changes inversely with volume, which is captured in the integral calculation. The result tells us the total work performed during the expansion process.
Gas Laws
The fundamental gas laws are essential in understanding the behavior of gases under various conditions of pressure, volume, and temperature. For adiabatic processes, we utilize a particular form of the gas law in our calculation, namely the relationship \( pv^{1.4} = c \).
This is a specific form of the ideal gas law modified for adiabatic processes. Each variable's relationship describes how any change in volume impacts pressure and vice versa without heat exchange. The exponent \( 1.4 \), known as the adiabatic index or \( \gamma \), represents the specific heat capacities' ratio for the gas in use.
Understanding these relationships helps predict how a gas behaves when subjected to changes, much like in our example. Here, knowing the volume quadruples helped in determining the final pressure using the established constant.
This is a specific form of the ideal gas law modified for adiabatic processes. Each variable's relationship describes how any change in volume impacts pressure and vice versa without heat exchange. The exponent \( 1.4 \), known as the adiabatic index or \( \gamma \), represents the specific heat capacities' ratio for the gas in use.
Understanding these relationships helps predict how a gas behaves when subjected to changes, much like in our example. Here, knowing the volume quadruples helped in determining the final pressure using the established constant.
Integral Calculus
Integral calculus is a mathematical tool used here to calculate the work done during the gas expansion. The work is represented by an integral of pressure over volume, as shown in the formula \( W = \int_{v_1}^{v_2} p \, dv \).
By substituting the specific relation for an adiabatic process, \( p = \frac{80}{v^{1.4}} \), the integral becomes a calculation of \( \int_{1}^{4} v^{-1.4} \, dv \).
To solve this, we apply the antiderivative of \( v^{-1.4} \), represented as:
By substituting the specific relation for an adiabatic process, \( p = \frac{80}{v^{1.4}} \), the integral becomes a calculation of \( \int_{1}^{4} v^{-1.4} \, dv \).
To solve this, we apply the antiderivative of \( v^{-1.4} \), represented as:
- \( F(v) = \frac{80}{-0.4} v^{-0.4} + C \)
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. In this context, we focus on the change in the energy of a system, and how this impacts the state variables, such as pressure and volume.
During an adiabatic expansion, the system is isolated from thermal exchange, and any energy transferred becomes work.In our problem, the key equation \( p v^{1.4} = c \) reflects the adiabatic condition, stipulating that energy changes manifest solely as work. This follows from the first and second laws of thermodynamics, where energy conservation and transfer through work or heat energy are central concepts. These laws guide calculations and predictions about the behavior of gases during transitions.
Understanding these principles helps explain how systems can do work without direct heat energy being exchanged, as demonstrated in the expansion process tackled in this exercise.
During an adiabatic expansion, the system is isolated from thermal exchange, and any energy transferred becomes work.In our problem, the key equation \( p v^{1.4} = c \) reflects the adiabatic condition, stipulating that energy changes manifest solely as work. This follows from the first and second laws of thermodynamics, where energy conservation and transfer through work or heat energy are central concepts. These laws guide calculations and predictions about the behavior of gases during transitions.
Understanding these principles helps explain how systems can do work without direct heat energy being exchanged, as demonstrated in the expansion process tackled in this exercise.
