Question

A piston confines 0.200 mol Ne(g) in 1.20 at 25 degree C. Two experiments are performed. (a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm. (b) The gas is allowed to expand reversibly and isothermally to the same final volume. Please calculate the work done by the gas system in these two processes, respectively. Which process does more work? (revised from 6/e exercise 8.11) Please show calculation details.

Answer 1

To calculate the work done by the gas system in two experiments, we can use different equations. The work done in experiment (a) is -1.20 atm·L, while the work done in experiment (b) is approximately -8.2931 atm·L. The work done in experiment (b) is greater than in experiment (a).

Step-by-step explanation:

To calculate the work done by a gas system, we can use the equation:

W = -PΔV

Where W is the work done, P is the pressure, and ΔV is the change in volume.

(a) For the first experiment, the gas expands against a constant pressure of 1.00 atm. The change in volume is 1.20 L. Plugging in these values, we get:

W = -1.00 atm * 1.20 L = -1.20 atm·L

(b) For the second experiment, the gas expands isothermally. We know that for an isothermal process, the work done by the gas is given by:

W = -nRT ln(Vf/Vi)

Where n is the number of moles, R is the ideal gas constant, T is the temperature, Vf is the final volume, and Vi is the initial volume. Plugging in the given values:

W = -0.200 mol * 0.0821 atm·L/mol·K * 298 K * ln(2.40 L / 1.20 L)

Calculating this expression gives:

W ≈ -8.2931 atm·L

Comparing the two values of work done, we find that the work done in experiment (b) is greater than in experiment (a).

Answer 2

The second experiment (reversible path) does more work

Step-by-step explanation:

Step 1:

A piston confines 0.200 mol Ne(g) in 1.20L at 25 degree °C

(a) The gas is allowed to expand through an additional 1.20 L against a constant of 1.00atm

Irreversible path: w =-Pex*ΔV

⇒ with Pex = 1.00 atm

⇒ with ΔV = 1.20 L

W = -(1.00 atm) * 1.20 L

W = -1.20L*atm *101.325 J /1 L*atm = -121.59 J

(b) The gas is allowed to expand reversibly and isothermally to the same final volume.

W = -nRTln(Vfinal/Vinitial)

⇒ with n = the number of moles = 0.200

⇒ with R = gas constant = 8.3145 J/K*mol

⇒ with T = 298 Kelvin

⇒ with Vfinal/Vinitial = 2.40/1.20 = 2

W = -(0.200mol) * 8.3145 J/K*mol *298K *ln(2.4/1.2)

W = -343.5 J

The second experiment (reversible path) does more work