Question

Thermodynamics:

Two experiments with a monatomic ideal gas were done. See data table for more information.

Initial conditions for both:

Volume:1.62L

Temperature: 600K

Pressure: 3.25 atm.

Was experiment performed in a one step process either isothermally or adiabatically? Or was it done in a combination of both isothermal and adiabatic steps? Please give calculated proof.

Could either experiments been done in a completely reversible manner?

Please also calculat the final conditions of volume and pressue of the gas for both experiments based on final temperature.

Data:

Experiment 1:

W = -3.83 * 10^2 J  

Q = 0

ΔS_sys (J/K) = 9.26 * 10^-2

Final temp: 313K

Experiment 2

W = -8.08 * 10^2 J  

Q = 3.80 * 10^2

ΔS_sys (J/K) = 6.33 * 10^-2

Final temp: 279K

Answer 1

EXPERIMENT 1:

Because heat is zero, the process was carried out in a adiabatic step. According to the relationship between temperature and volume for adiabatic process:

Substituting R = 8.314 J/mol*K ; n = 1 mol and Cv = 3/2 R for an ideal monoatomic gas, and solving for V_2^:

Applying the known relationship pressure-volume for adiabatic process:

Solving for P₂:

Substituting known values:

EXPERIMENT 2:

Because heat is not zero, and temperature is not constant, we assume that the process is carried out with an isothermical step followed by an adiabatic step. In the isothermal process, we have:

Substituting known data for heat:

Solving for V₂:

Applying the Boyle's law:

Solving for P₂:

V₂ and P₂ are the initial conditions for the adiabatic step. Applying the known relationship temperature-volume for adiabatic processes:

Substituting known values, and solving for V₃:

Applying the known relationship pressure volume for adiabatic process:

Solving for P₃: