Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Consider a gas in a closed system that follows a reversible isothermal path from

ID: 1718098 • Letter: C

Question

Consider a gas in a closed system that follows a reversible isothermal path from an initial state (p_1, T) to a final state (p_2, T). If T = 300 K, p_1 = 10 bar, and p_2 = 1 bar, answer the following questions and provide numerical answers with correct units. If the gas is an ideal gas, Calculate the work along the path; is work done on or by the system? calculate DeltaU along the Path Consider a non-isothermal path between the same initial and same final states. What is DeltaU for this second path? How much work is done along the second path? Explain. Suppose instead that the gas in question follows the same isothermal path as the ideal gas but that the gas behaves according to the truncated virial equation of state given by z = 1 + Brho + Crho^2. If the second and third virial coefficients for the gas are B = -121.7 cm^3/mol and C = 4927 cm^6/mol^2, answer the following Calculate the molar density, rho, of the virial gas at the initial and final state. What is the work along path? calculate DeltaU along the path. For a non-isothermal path between the same initial and same final states, what is DeltaU? How much work is done along the second path? Explain. Suppose now the virial gas follows a reversible adiabatic path between the same initial and final states. What is the amount of work along the path? What is DeltaU for the adiabatic path?

Explanation / Answer

Given data

Temperature = T=300K

p1=10bar

p2=1bar

Requirement

Work done along this path = W=?

     Solution

As the process is isothermal and we are given with pressure and temperature so Boyles law is applicable. B y ist law of thermodynamics

   dU = Q - W

   W = Q- dU

As process is isothermal , thats mean temperature will remain constant , so molecules will have no change in their kinetic energy because kinetic energy of the molecules is the function of temperature. Therefore dU= 0

    W = Q

    W =Q= nRT ln P1 / P2

As path is the same so we can find work by our initial data.

     W = (1mol)(8.13 J / K mol)ln (10bar / 1bar)

     W = 18.72J

This would be the work done along this path.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote