Original volume of an ideal gas is 2x10−3 m3 at pressure value of 1.2x105 N m2 and at temperature of 0oC. The gas takes part in the thermodynamic cycle given here: As first step the pressure of the gas is increased to 1.5x105 Pa keeping the volume constant. Next step is to expand the gas isothermally to the original pressure value. After it, as last process is made: the gas is gone back to its original status keeping the pressure fixed. What is the temperature during the isothermal process? What is the volume reached at the end of the isothermal process? What is the work done by the gas during its whole cycle?

To tackle this problem, we need to analyze the thermodynamic cycle of the ideal gas step by step. We’ll use the ideal gas law and the principles of thermodynamics to find the temperature during the isothermal process, the volume at the end of the isothermal expansion, and the total work done by the gas throughout the cycle.

Step 1: Understanding the Initial Conditions

The initial conditions of the gas are as follows:

• Initial Volume (V₁): 2 x 10^-3 m³
• Initial Pressure (P₁): 1.2 x 10⁵ N/m²
• Initial Temperature (T₁): 0°C = 273.15 K

Using the ideal gas law, which states that PV = nRT, we can find the number of moles (n) of the gas:

Calculating the Number of Moles

Rearranging the ideal gas law gives us:

n = PV / RT

Substituting the known values:

n = (1.2 x 10⁵ N/m²) * (2 x 10^-3 m³) / (8.314 J/(mol·K) * 273.15 K)

Calculating this yields:

n ≈ 0.088 moles

Step 2: The Isothermal Expansion

In the first step of the cycle, the pressure is increased to 1.5 x 10⁵ Pa while keeping the volume constant. This step does not change the volume, so we can denote the new state as:

• New Pressure (P₂): 1.5 x 10⁵ N/m²
• Volume remains V₁: 2 x 10^-3 m³

Next, the gas expands isothermally back to the original pressure (P₁ = 1.2 x 10⁵ Pa). During this process, the temperature remains constant, and we need to find this temperature (T_iso).

Finding the Temperature During Isothermal Process

Using the ideal gas law again for the isothermal process:

P₂V₂ = nRT_iso

Since we know the pressure (P₂ = 1.5 x 10⁵ Pa) and the volume remains constant (V₂ = V₁), we can express the temperature:

T_iso = P₂V₁ / (nR)

Substituting the values:

T_iso = (1.5 x 10⁵ N/m²) * (2 x 10^-3 m³) / (0.088 mol * 8.314 J/(mol·K))

Calculating this gives:

T_iso ≈ 1021.5 K

Step 3: Volume at the End of Isothermal Process

During the isothermal expansion, we need to find the volume (V₂) when the gas returns to the original pressure:

Using the ideal gas law again:

P₁V₂ = nRT_iso

Rearranging gives:

V₂ = nRT_iso / P₁

Substituting the values:

V₂ = (0.088 mol * 8.314 J/(mol·K) * 1021.5 K) / (1.2 x 10⁵ N/m²)

Calculating this yields:

V₂ ≈ 0.0006 m³ or 6 x 10^-4 m³

Step 4: Work Done by the Gas During the Cycle

To find the total work done by the gas during the entire cycle, we need to consider the work done during the isothermal expansion:

Work (W) during isothermal expansion is given by:

W = nRT_iso ln(V₂/V₁)

Substituting the known values:

W = 0.088 mol * 8.314 J/(mol·K) * 1021.5 K * ln(6 x 10^-4 m³ / 2 x 10^-3 m³)

Calculating this gives:

W ≈ - 0.088 * 8.314 * 1021.5 * ln(0.3) ≈ - 0.088 * 8.314 * 1021.5 * (-1.20397) ≈ 0.088 * 8.314 * 1021.5 * 1.20397

W ≈ 0.088 * 8.314 * 1223.5 ≈ 0.88 kJ

Summary of Results

• Temperature during the isothermal process: T_iso ≈ 1021.5 K
• Volume at the end of the isothermal process: V₂