Question:

A gas undergoes a thermodynamic process described by the following steps:

1. The gas is initially at a temperature of $T_i = 300 \, \text{K}$ and a pressure of $P_i = 3 \, \text{atm}$.
2. The gas is compressed adiabatically to one-third of its initial volume.
3. The gas is then cooled at constant volume until its temperature reaches $T_f = 100 \, \text{K}$.

Given that the molar specific heat capacity of the gas at constant volume is $C_v = 2.5 \, \text{cal/mol K}$, and the molar gas constant is $R = 8.314 \, \text{J/mol K}$, determine:

a) The final pressure $P_f$ of the gas after the adiabatic compression.

b) The work done on or by the gas during the adiabatic compression.

c) The heat transferred during the cooling process from step 2 to step 3.

Answer:

a) To find the final pressure $P_f$ after the adiabatic compression, we can utilize the relation for an adiabatic process:

where $V_i$ and $V_f$ are the initial and final volumes, and $\gamma$ is the heat capacity ratio which for a monoatomic ideal gas is $\gamma = \frac{C_p}{C_v} = \frac{5}{2}$.

Since the gas is compressed to one-third of its initial volume, the final volume is $V_f = \frac{1}{3}V_i$. Substituting these values into equation (1), we have:

Simplifying equation (2), we get:

Evaluating the expression, we find:

Therefore, the final pressure after the adiabatic compression is approximately $0.512 \, \text{atm}$.

b) The work done on or by the gas during an adiabatic process can be calculated using the following relation:

We already know the values of $P_i$, $V_i$, $P_f$, and $V_f$. Substituting these values into equation (6), we have:

Therefore, the work done on or by the gas during the adiabatic compression is $\frac{8}{9}V_i$.

c) During the constant volume cooling from step 2 to step 3, the heat transferred can be calculated using the first law of thermodynamics:

Since the process occurs at constant volume, there is no change in volume ($\Delta V = 0$), and therefore the work done by the gas is zero ($W = 0$).

The change in internal energy $\Delta U$ can be calculated using the equation:

where $n$ is the number of moles of the gas, $C_v$ is the molar specific heat capacity at constant volume, and $\Delta T$ is the change in temperature.

The number of moles $n$ can be calculated using the ideal gas equation:

Given that the initial pressure $P_i = 3 \, \text{atm}$ and the initial temperature $T_i = 300 \, \text{K}$, we can rearrange equation (12) to solve for $n$:

Substituting the known values into equation (13):

Simplifying equation (14), we find:

Now we can calculate $\Delta U$ using equation (11), and since $Q = \Delta U + W$, the heat transferred is equal to $\Delta U$:

Converting the units from cal to J, we have:

Therefore, the heat transferred during the cooling process from step 2 to step 3 is approximately $-2.109 \cdot V_i \, \text{J}$. Note that the negative sign indicates heat is transferred from the gas to the surroundings.