Ideal nonreacting gases A and B are contained inside a perfectly insulated chamber, separated by a thin partition, as shown in the figure. The partition is removed, and the two gases mix till final equilibrium is reached. The change in total entropy for the process is _________J/K (rounded off to 1 decimal place).
Given: Universal gas constant \( R = 8.314 \) J/(mol K), \( T_A = T_B = 273 \) K, \( P_A = P_B = 1 \) atm, \( V_B = 22.4 \) L, \( V_A = 3V_B \).
Ideal nonreacting gases A and B are contained inside a perfectly insulated chamber, separated by a thin partition, as shown in the figure. The partition is removed, and the two gases mix till final equilibrium is reached. The change in total entropy for the process is _________J/K (rounded off to 1 decimal place).
Given: Universal gas constant \( R = 8.314 \) J/(mol K), \( T_A = T_B = 273 \) K, \( P_A = P_B = 1 \) atm, \( V_B = 22.4 \) L, \( V_A = 3V_B \).
Show Hint
Correct Answer: 18.5
Solution and Explanation
In this setup, the two gases are mixed in an adiabatic and isothermal process. Given the conditions, we can calculate the change in entropy.
Step 1: Calculate the initial and final volumes and the number of moles:
Volume of Gas A \( V_A = 3 \times 22.4 \, L = 67.2 \, L \)
Volume of Gas B \( V_B = 22.4 \, L \)
Total final volume \( V_{final} = V_A + V_B = 89.6 \, L \)
Step 2: Calculate the initial number of moles for each gas using \( PV = nRT \):
Number of moles of Gas A \( n_A = \frac{P \times V_A}{R \times T} \)
Number of moles of Gas B \( n_B = \frac{P \times V_B}{R \times T} \)
Step 3: Calculate the change in entropy for each gas: \[ \Delta S_A = n_A \times R \times \ln\left(\frac{V_{final}}{V_A}\right) \] \[ \Delta S_B = n_B \times R \times \ln\left(\frac{V_{final}}{V_B}\right) \] Step 4: Sum the changes in entropy: \[ \Delta S = \Delta S_A + \Delta S_B \] The change in total entropy for the process is \( 18.7 \, J/K \) rounded to 1 decimal place.
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