An ideal gas in a moveable piston is allowed to reversibly expand by slowly heating at constant
pressure.
a. Derive the equation for the work done reversibly as a function of the initial and final temperatures.( i have used w=-nrtln(v final/v intial) i dont know how can I take derivative.
b. Derive the equation for the change in heat in the system which is q.
c. Derive the equation for the change in entropy.
I am not sure for b and c.
a. To derive the equation for the work done, we need to consider the process of reversible expansion of the gas in the movable piston. In this case, the pressure remains constant throughout the process.
We can use the equation for work done by a gas, which is given by:
w = -P∆V
where w is the work done, P is the pressure, and ∆V is the change in volume.
Since the pressure remains constant, we can substitute P with a constant value, let's say P1.
w = -P1∆V
To relate the change in volume (∆V) to the initial and final temperatures, we can use the ideal gas law:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
By rearranging the ideal gas law, we can express the initial volume (V1) and final volume (V2) in terms of the initial temperature (T1) and final temperature (T2):
V1 = (nRT1) / P1
V2 = (nRT2) / P1
Substituting these expressions into the equation for work done, we have:
w = -P1(V2 - V1)
w = -P1[((nRT2) / P1) - ((nRT1) / P1)]
w = -nR(T2 - T1)
Therefore, the equation for the work done reversibly as a function of the initial and final temperatures is:
w = -nR(T2 - T1)
b. To derive the equation for the change in heat (q) in the system, we can use the first law of thermodynamics:
∆U = q - w
where ∆U is the change in internal energy, q is the heat added to the system, and w is the work done on the system.
For a reversible, constant pressure process, the change in internal energy (∆U) can be expressed as:
∆U = nCp(T2 - T1)
where Cp is the heat capacity at constant pressure.
By rearranging the equation, we can solve for q:
q = ∆U + w
q = nCp(T2 - T1) + (-nR(T2 - T1))
q = n(Cp - R)(T2 - T1)
Therefore, the equation for the change in heat in the system is:
q = n(Cp - R)(T2 - T1)
c. To derive the equation for the change in entropy, we need to use the definition of entropy:
∆S = q / T
From part b, we know that q = n(Cp - R)(T2 - T1). Substituting this into the equation for ∆S, we have:
∆S = (n(Cp - R)(T2 - T1)) / T
∆S = n(Cp - R) / T * (T2 - T1)
Therefore, the equation for the change in entropy is:
∆S = n(Cp - R) / T * (T2 - T1)