One mole of an ideal monatomic gas initially at 300 K is expanded from an initial pressure of

10 atm to a final pressure of 1 atm. Calculate ��E, q, w, ��H, and the final temperature T2 for
this expansion carried out according to each of the following paths. The molar heat capacity
at constant volume for a monatomic gas is CV = 3/2 nR. Taken from Tinoco 2.8.
(a) An isothermal, reversible expansion.
(b) An expansion against a constant external pressure of 1 atm in a thermally isolated
(adiabatic) system.
Hint for part b: The transition is not isothermal. Be patient and write down everything you
know. You will find that ��E = -1.35 x 103 J.
(c) An expansion against zero external pressure (against a vacuum) in an adiabatic system.

a)

dU = 0 J
dH = 0 J
w = -5743.11 J
q = 5732.11 J

b)
q = 0 J
dU = -1350 J
w = -1350 J
V2 = 15.78 L
T2 = 192.3 K
dH = -2238.55 J

To calculate the values for each of these paths, we need to use the first law of thermodynamics, which states that the change in internal energy (��E) of a system is equal to the heat (q) added to the system minus the work (w) done by the system:

��E = q - w

We can also use the equation for the molar heat capacity at constant volume (CV) for a monatomic gas:

CV = 3/2 nR

where n is the number of moles of the gas and R is the ideal gas constant.

Now let's calculate the values for each path:

(a) Isothermal, reversible expansion:
For an isothermal process, the temperature remains constant (T1 = T2). Since the initial and final temperatures are the same, there is no change in internal energy (��E = 0).

Using the equation for an isothermal process, we know that q = -w. So, to calculate q, we can use:

q = -w = -(P2V2 - P1V1)

where P1 and P2 are the initial and final pressures, and V1 and V2 are the initial and final volumes.

To find the final volume, we can use the ideal gas law:

P1V1 / T1 = P2V2 / T2

Solving for V2:

V2 = (P1V1 * T2) / (P2 * T1)

Since the gas is ideal, we can use the ideal gas law to relate the volume and the number of moles:

V1 / n = V2 / n

Simplifying the equation:

V1 = V2

So, the equation becomes:

V2 / n = (P1V1 * T2) / (P2 * T1)

Now, solving for T2:

T2 = (P2 * T1 * V2) / (P1 * V1)

(b) Expansion against a constant external pressure of 1 atm in a thermally isolated (adiabatic) system:
In an adiabatic process, no heat is exchanged between the system and the surroundings (q = 0). Therefore, the change in internal energy is equal to the work done by the system:

��E = -w

To calculate the work done, we can use the equation:

w = -Pext * ∇V

where Pext is the external pressure and ∇V is the change in volume (V2 - V1).

Since the process is adiabatic and the system is thermally isolated, we know that q = 0, so the change in internal energy is:

��E = q - w = 0 - (-Pext * ∇V) = Pext * ∇V

The work done by the system is given as -1.35 x 10^3 J, therefore:

Pext * ∇V = -1.35 x 10^3 J

(c) Expansion against zero external pressure (against a vacuum) in an adiabatic system:
Since the external pressure is zero, no work is done against it (w = 0). Therefore, the change in internal energy is equal to the heat added:

��E = q

However, since the process is adiabatic, there is no heat exchange (q = 0). Therefore, the change in internal energy is also zero (��E = 0).

To summarize:
(a) An isothermal, reversible expansion:
��E = 0
q = -w
T2 = (P2 * T1 * V2) / (P1 * V1)

(b) An expansion against a constant external pressure of 1 atm in a thermally isolated (adiabatic) system:
��E = -w = -1.35 x 10^3 J
q = 0

(c) An expansion against zero external pressure (against a vacuum) in an adiabatic system:
��E = 0
q = 0
w = 0

You can plug in the values for the initial pressure, final pressure, initial temperature, and initial volume to calculate the final temperature and other relevant quantities for each path.