Let's break down the problem **step by step**: --- ## **Step 1: Understanding the Problem** - **Two compartments (A and B)**, separated by a barrier. - **Chemical potential on A** is **1.5 times** higher than B. - **Temperatures**: - \(T_A = 303\,K\) - \(T_B = 363\,K\) - The barrier is replaced with a **semipermeable membrane**: particles and water can now move freely. - **At equilibrium**, what is the ratio \(\frac{N_A}{N_B}\) (number of particles in A to B)? --- ## **Step 2: Key Concept — Chemical Potential Equality at Equilibrium** When exchange is allowed, **at equilibrium**: \[ \mu_A(T_A, N_A) = \mu_B(T_B, N_B) \] --- ## **Step 3: Express the Chemical Potential** For an ideal solution: \[ \mu = \mu^ + k_B T \ln N \] where \(N\) is the number of particles (proportional to concentration). Let’s denote the chemical potentials before membrane as: \[ \mu_A^\text{initial} = 1.5 \mu_B^\text{initial} \] But after the membrane is in place and equilibrium is reached, the chemical potential in both compartments must be equal. --- ## **Step 4: Set Up the Equation** At equilibrium: \[ \mu_A(T_A, N_A) = \mu_B(T_B, N_B) \] \[ \mu^ + k_B T_A \ln N_A = \mu^ + k_B T_B \ln N_B \] \[ k_B T_A \ln N_A = k_B T_B \ln N_B \] \[ T_A \ln N_A = T_B \ln N_B \] \[ \ln N_A = \frac{T_B}{T_A} \ln N_B \] \[ \ln N_A = 1.198 \ln N_B \] \[ \ln N_A - 1.198 \ln N_B = \] \[ \ln \left( \frac{N_A}{N_B^{1.198}} \right) = \] \[ \frac{N_A}{N_B^{1.198}} = 1 \] \[ N_A = N_B^{1.198} \] \[ \frac{N_A}{N_B} = N_B^{.198} \] But, we need a number, not in terms of \(N_B\). --- ## **Step 5: Relate Initial Chemical Potentials** Before the membrane: \[ \mu_A = 1.5 \mu_B \] \[ \mu^ + k_B T_A \ln N_A^\text{initial} = 1.5 (\mu^ + k_B T_B \ln N_B^\text{initial}) \] But at equilibrium, the chemical potentials are equalized, **regardless of initial numbers**, so the equilibrium ratio is **independent of initial conditions**. --- ## **Step 6: Final Ratio Calculation** From above: \[ T_A \ln N_A = T_B \ln N_B \] \[ \ln N_A = \frac{T_B}{T_A} \ln N_B \] \[ \ln N_A = 1.198 \ln N_B \] \[ N_A = N_B^{1.198} \] \[ \frac{N_A}{N_B} = N_B^{.198} \] But without a given \(N_B\), the only possible answer is to express the ratio in terms of \(N_B\). But, typically, for such questions, we compare concentrations, so let's rewrite as follows. Assume the **concentration** in each compartment at equilibrium is: \[ \frac{N_A}{N_B} = e^{\left(\frac{T_B-T_A}{T_A}\ln N_B\right)} \] But since there is no other information (volumes, total particles), the ratio at equilibrium is: \[ T_A \ln N_A = T_B \ln N_B \implies \ln N_A = \frac{T_B}{T_A} \ln N_B \implies N_A = N_B^{T_B/T_A} \] Given: - \(T_A = 303\) - \(T_B = 363\) So, \[ \frac{T_B}{T_A} = \frac{363}{303} = 1.198 \] \[ N_A = N_B^{1.198} \] \[ \frac{N_A}{N_B} = N_B^{.198} \] But if you want the **ratio of concentrations** at equilibrium for arbitrary \(N_B\), you can only express it in terms of \(N_B\). Usually, unless a specific value is given, the ratio is: \[ \boxed{\frac{N_A}{N_B} = N_B^{.198}} \] But if we assume **equal volumes and total particles are conserved**, the answer remains as above. If you have any specific values, substitute for \(N_B\). --- ## **Final Boxed Answer** \[ \boxed{\frac{N_A}{N_B} = N_B^{.198}} \] Or, if you assume **equal initial concentrations** and want a numerical value, you can say: \[ \frac{N_A}{N_B} = e^{(T_B/T_A-1)\ln N_B} = N_B^{.198} \] If a specific \(N_B\) is given, substitute it here. --- ### **Summary** - At equilibrium, chemical potentials are equal. - The ratio depends on the temperature ratio. - If you need a number, you need \(N_B\). Otherwise, write the answer in terms of \(N_B\). **Let me know if you have a specific value for \(N_B\)!**