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, \( V_A = 3V_B \).
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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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