Question

An air cannon uses compressed air to propel a projectile from rest to a final velocity. Consider an air cannon that is to accelerate a 10 -gram projectile to a speed of \(300 \mathrm{m} / \mathrm{s}\) using compressed air, whose temperature cannot exceed \(20^{\circ} \mathrm{C}\) The volume of the storage tank is not to exceed \(0.1 \mathrm{m}^{3} .\) Select the storage volume size and maximum storage pressure that requires the minimum amount of energy to fill the tank.

Short answer

Answer: The storage volume size that requires the minimum amount of energy to fill the tank is 0.1 m^3, and the maximum storage pressure is approximately 105625 Pa.

Step-by-step solution

Recall Key Equations

In this problem, we need to use the ideal gas law and the conservation of energy. The ideal gas law equation is \(PV=nRT\), where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin. The conservation of energy equation is \(W = \dfrac{1}{2}mv^2\), where W is the work done, m is the mass of the projectile, and v is the final velocity.

Convert Units to SI and Temperature to Kelvin

Convert the mass of the projectile to kilograms and the temperature to Kelvin: m = 10 grams = 0.01 kg, T = 20°C = 293.15 K.

Calculate the Work Done and Number of Moles of Gas

Calculate the work done by substituting the given values into the conservation of energy equation: W = (1/2)(0.01 kg)(300 m/s)^2 = 450 J. Then, solve for the number of moles (n) using the ideal gas law, assuming standard pressure of 101325 Pa for the tank when empty: n = (101325 Pa)(0.1 m^3) / (8.314 J/(mol·K)·293.15 K) ≈ 4.127 mol.

Determine the Final Pressure

We can calculate the final pressure using the work-energy theorem: W = Δ(PV). Since P_initial * V_initial = P_final * V_final and V_initial = V_final, that means that P_final = P_initial + ΔP or ΔP = P_final - P_initial. So, the work-energy theorem becomes W = V * (P_final - P_initial). Rearranging for P_final, we get P_final = P_initial + (W/V). Substituting known values, we have P_final = 101325 Pa + (450 J / 0.1 m^3) ≈ 105625 Pa.

Calculate the Minimum Energy Required

Now we need to minimize the energy required to fill the tank. The energy required is determined by the work done on the gas to increase its pressure. As the volume increases, the amount of work required to compress the gas to achieve the same pressure decreases. Thus, to minimize the energy required, we should use the maximum allowable volume of 0.1 m^3.

Provide the Results

The storage volume size that requires the minimum amount of energy to fill the tank is 0.1 m^3, and the maximum storage pressure is approximately 105625 Pa.