Problem 18

Question

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.

Step-by-Step Solution

Verified

Answer

The work done is approximately 180.46 foot-pounds.

1Step 1: Identify Known Variables

We start by identifying the known variables from the problem statement. The initial volume $V_1$ is 1 cubic foot, the initial pressure $P_1$ is 80 pounds per square inch, the final volume $V_2$ is 4 cubic feet, and the law governing the expansion is $p v^{1.4} = c$. This means the process is adiabatic.

2Step 2: Apply Adiabatic Law to Find Constant

The adiabatic law states $p v^{1.4} = c$. For the initial state, $c = P_1 V_1^{1.4}$. Substituting the known values, $c = 80 \times 1^{1.4} = 80$.

3Step 3: Express Final Pressure Using Adiabatic Law

Using the adiabatic law $P_2 V_2^{1.4} = c$, substitute $c = 80$ and $V_2 = 4$ to solve for $P_2$. Thus, $P_2 = \frac{80}{4^{1.4}} = \frac{80}{10.24}\approx 7.81\text{ psi}$.

4Step 4: Setup Integral for Work Done

The work done during adiabatic expansion from $V_1$ to $V_2$ is calculated using the integral $W = \int_{V_1}^{V_2} P \, dV$. Since $P = \frac{c}{V^{1.4}}$, we substitute into the integral: \[W = \int_{1}^{4} \frac{80}{V^{1.4}} \, dV.\]

5Step 5: Solve the Integral

Evaluate the integral:\[W = 80 \int_{1}^{4} V^{-1.4} \, dV.\]The integral of $V^{-1.4}$ is $\frac{V^{-0.4}}{-0.4} = -2.5 V^{-0.4}$. Evaluate it from 1 to 4:\[W = 80 \left[-2.5 \left(V^{-0.4}\right)\right]_{1}^{4} = 80 \left[-2.5 \left(4^{-0.4} - 1^{-0.4}\right)\right] = 80 \times 2.5 \left(1 - \frac{1}{10.24}\right).\]Calculate $80 \times 2.5 \times 0.9023 \approx 180.46\text{ ft-lbs}$.

6Step 6: Final Answer

The work done by the gas during the adiabatic expansion from 1 cubic foot to 4 cubic feet at 80 psi initial pressure is approximately 180.46 foot-pounds.

Key Concepts

Work Done by a GasAdiabatic ProcessCalculus Integration

Work Done by a Gas

When a gas expands, it does work against external pressure. This work can be calculated by examining changes in pressure and volume during the process. In scenarios like adiabatic expansion, where no heat is exchanged with the surroundings, this becomes especially significant.
The formula to find the work done ($ W $ ) by a gas is given by:

\[ W = \int_{V_1}^{V_2} P \, dV \]

Here, $ P $ is the pressure as a function of volume $ V $, and the integral is evaluated from the initial volume $ V_1 $ to the final volume $ V_2 $.
In this problem, the expansion rule is governed by the relationship $ p v^{1.4} = c $, known as an adiabatic process. Applying this law allows us to express pressure $ P $ as a function of volume so that the integration for work can proceed. Understanding how these variables interrelate is key to accurately finding the work done by the gas.

Adiabatic Process

An adiabatic process is a thermodynamic sequence where no heat is exchanged with the surroundings. This means all the energy transfer in or out of the system is done in the form of work. For gases undergoing adiabatic processes, pressure, volume, and temperature change according to specific rules.
In this context, the rule is given by the adiabatic equation:

\[ p v^{\gamma} = c \]

where $ p $ is the pressure, $ v $ is the volume, and $ \gamma $ (gamma) is the heat capacity ratio, specific to the gas, often 1.4 for diatomic gases like air.
As the gas expands adiabatically, pressure decreases more rapidly compared to isothermal expansion because the internal energy and thus the temperature of the gas decreases. This is crucial for calculating the integrals required for determining the work done, as it involves expressing pressure as a function of a changing volume.

Calculus Integration

Calculus integration is a mathematical technique used to find the area under a curve. In the context of physics, it helps us calculate total quantities, like work, when the conditions are constantly changing.
To calculate the work done by a gas during expansion, we set up an integral of pressure over changing volume:

\[ W = \int_{V_1}^{V_2} \frac{c}{V^{1.4}} \, dV \]

In our specific case, using the values $ c = 80 $, $ V_1 = 1 $, and $ V_2 = 4 $, the function becomes:

\[ W = 80 \int_{1}^{4} V^{-1.4} \, dV \]

Integration techniques require understanding how to find antiderivatives. For a function $ V^{n} $, the antiderivative is $ \frac{V^{n+1}}{n+1} $, and in our case, this gives us: