Question

One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown below. If the work done by the gas in the two processes are the same, the value of \(\ln \frac{v_3}{v_2}\) is _____

(U: internal energy, S: entropy, p: pressure, V: volume, R: gas constant)
(Given: molar heat capacity at constant volume, \(C_v,m\) of the gas is \(\frac{5}{2}\)R)

Answer 1

Correct Answer: 10

Step 1: Process I – Adiabatic Reversible Process:

For an adiabatic reversible process, the first law of thermodynamics tells us that the change in internal energy \( \Delta U \) is equal to the work done by the system since no heat is exchanged with the surroundings.
Given the equation \( \frac{\Delta U}{R} = 450 - 2250 \), we can calculate the change in internal energy:
\(\Delta U = -1800R\).
Since the process is adiabatic, the work done \( W_I \) is equal to the change in internal energy:
\( W_I = \Delta U = -1800R \).

Step 2: Process II – Reversible Isothermal Process:

In the second process, the temperature \( T_1 \) is given as 900 K.
Since the process is isothermal, the temperature remains constant during the process.
We are tasked with finding the temperature \( T_2 \) after the reversible adiabatic process.

Step 3: Calculation of \( T_2 \) after the Reversible Adiabatic Process:

For an adiabatic process, the relation between the change in temperature and internal energy is given by:
\(-1800R = nC_v(T_2 - T_1)\),
where:
\( n \) is the number of moles, \( C_v \) is the specific heat capacity at constant volume, \( T_1 \) is the initial temperature, and \( T_2 \) is the final temperature.
For helium, we know that \( C_v = \frac{5}{2}R \). Now, substitute the given values into the equation:
\(-1800R = n \times \frac{5}{2}R \times (T_2 - 900)\).
Canceling out \( R \) from both sides and solving for \( T_2 \):
\(-1800 = n \times \frac{5}{2} \times (T_2 - 900)\).
Substitute \( n = 1 \) (assuming one mole of helium), and simplify the equation:
\(-1800 = \frac{5}{2} \times (T_2 - 900)\).
Multiplying both sides by 2:
\(-3600 = 5(T_2 - 900)\).
Now, divide by 5:
\(-720 = T_2 - 900\).
Finally, solving for \( T_2 \):
\( T_2 = 180 \, \text{K} \).

Step 4: Work Done in Process II:

The work done in a reversible isothermal process is given by:
\( W_{II} = -nRT_2 \ln \left( \frac{v_3}{v_2} \right) \),
where:
\( T_2 = 180 \, \text{K} \) and \( \ln \frac{v_3}{v_2} = 10 \).
Substitute these values into the equation:
\( W_{II} = -1 \times R \times 180 \times 10 \).
Thus, the work done in process II is:
\( W_{II} = -1800R \).

Final Answer:
The correct answer is: \( \ln \left( \frac{v_3}{v_2} \right) = 10 \).