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.
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