Question

\(1-1 \mathrm{bm}\) of air at 10 psia and \(70^{\circ} \mathrm{F}\) is contained in a piston-cylinder device. Next, the air is compressed reversibly to 100 psia while the temperature is maintained constant. Determine the total amount of heat transferred to the air during this compression.

Short answer

Answer: -95,175.8 J (The heat is removed from the air during the compression process).

Step-by-step solution

List the given information and assumptions.

Given information: - Initial pressure (P1) = 10 psia - Final pressure (P2) = 100 psia - Constant temperature (T) = 70°F Assumptions: - The air behaves as an ideal gas. - The compression is isothermal and reversible.

Convert units to SI.

We need to convert the given pressures and temperature from their current units to their SI units. There are several conversion factors, which include: 1 psi = 6894.76 Pa 1°F = 5/9(°C - 32) Converting the pressure and temperature values: P1 = 10 psia * (6894.76 Pa/1 psi) = 68,947.6 Pa P2 = 100 psia * (6894.76 Pa/1 psi) = 689,476 Pa T = (5/9) * (70°F - 32°F) = 21.11°C = 294.26 K Now, we have: P1 = 68,947.6 Pa P2 = 689,476 Pa T = 294.26 K

Calculate the initial and final volumes.

We are given that we have 1 bm (1 standard cubic meter) of air. Let's assume that the initial volume is V1, and the final volume is V2. Since the process is isothermal, the ideal gas law for the initial and final states can be written as: P1 * V1 = n * R * T P2 * V2 = n * R * T Since the number of moles (n) and the gas constant (R) are the same in both equations, we can write the relationship between P1, V1, P2, and V2 as: P1 * V1 = P2 * V2 From the given information, V1 = 1 m³. Therefore, we can solve for V2: V2 = (P1 * V1) / P2 = (68,947.6 Pa * 1 m³) / 689,476 Pa = 0.1 m³

Compute the work done during the compression.

As the compression is isothermal and reversible, the work done on the gas during the compression can be determined by the following equation: W = n * R * T * ln(V2/V1) We have P1 * V1 = n * R * T, which implies n * R = P1 * V1 / T. Substituting in the equation for work: W = (P1 * V1 / T) * T * ln(V2/V1) = P1 * V1 * ln(V2/V1) W = (68,947.6 Pa * 1 m³) * ln(0.1 m³/1 m³) = -95,175.8 J Note: Since the work value is negative, it implies that work is done on the gas during compression (as expected).

Determine the heat transfer.

For an isothermal process, the change in internal energy is zero. Thus, the first law of thermodynamics can be written as: ΔU = Q - W Since ΔU = 0, we can write: Q = W Therefore, the total amount of heat transferred to the air during this isothermal compression process is: Q = -95,175.8 J The heat transferred to the air is 95,175.8 J in the negative direction, which implies that the heat is actually being removed from the air during the compression process.