Question:

A gas sample undergoes a thermodynamic process described by the pressure-volume (PV) diagram shown below:

The gas initially occupies a volume of 2 m³ at a pressure of 3 atm. It expands isothermally to a final volume of 6 m³. The temperature of the gas is 300 K during the entire process.

a) Calculate the work done by the gas during this isothermal expansion.

b) Calculate the heat absorbed by the gas during this isothermal expansion.

c) Determine the change in internal energy of the gas.

d) Verify if the process is an endothermic or exothermic process.

Answer:

a) To calculate the work done by the gas, we use the formula:

W = -∫P dV

Since the process is isothermal, the equation of state for an ideal gas can be used, which is: PV = nRT

Rearranging the equation, we have: P = (nRT) / V

Let's define the following variables:

P1 = Initial pressure = 3 atm V1 = Initial volume = 2 m³ V2 = Final volume = 6 m³ T = Temperature = 300 K R = Ideal gas constant = 8.314 J/(mol·K)

Substituting the values into the equation of state, we can find the value of n (the number of moles):

P1 * V1 = nRT

(3 atm) * (2 m³) = n * (8.314 J/(mol·K)) * (300 K) 6 atm*m³ = 2.4942 J/mol

n ≈ 2.42 mol

Next, let's calculate the work done by the gas:

W = -∫(P dV)

We integrate with respect to volume, so:

W = -∫[(nRT) / V] dV

Integrating from V1 to V2:

W = -nRT ∫(1 / V) dV

W = -nRT * ln(V)|[V1, V2)

W = -nRT * [ln(V2) - ln(V1)]

W = -nRT * ln(V2 / V1)

Substituting the known values:

W = -(2.42 mol) * (8.314 J/(mol·K)) * (300 K) * ln(6 m³ / 2 m³)

W ≈ -7256 J

Therefore, the work done by the gas during this isothermal expansion is approximately -7.26 kJ.

b) Since the process is isothermal, the internal energy (ΔU) remains constant. Therefore, the heat absorbed (Q) is equal to the work done (W) by the gas:

Q = -7.26 kJ

Therefore, the heat absorbed by the gas during this isothermal expansion is approximately -7.26 kJ.

c) The change in internal energy (ΔU) of the gas is zero because the process is isothermal, and the internal energy remains constant.

ΔU = 0

Therefore, the change in internal energy of the gas is zero.

d) The process is considered an exothermic process because work is done by the gas (-7.26 kJ) and heat is released by the gas (-7.26 kJ).