Question

The above apparatus consists of three temperature jacketed 12.315 litre bulbs connected by stop cocks. Bulbs 'A' consist of $NH_{3(g)}$, $CO_{2(g)}$ & $N_{2(g)}$ at $27°C$ and a total pressure of 60 atm. Bulb 'B' and 'C' are empty and maintained at temperature of $-73°C$ and $-173°C$ respectively. Initially stop-cocks are closed and volume of the connecting lines is zero.

Given:

(i) $NH_{3(g)}$ liquifies at $-20°C$, $CO_{2(g)}$ liquifies at $-80°C$ and $N_{2(g)}$ liquifies at $-180°C$.

(ii) Vapour pressure of $NH_{3(l)}$ at $-73°C$ = 1 atm

(iii) Vapour pressure of $NH_{3(l)}$ at $-173°C$ is negligible

(iv) Vapour pressure of $CO_{2(l)}$ is negligible

(v) On opening stop-cock (i) only, the pressure in containers becomes 17 atm and, on opening both stop-cocks pressure decreases to 4 atm

what is the value of $(n_{NH_3}+n_{CO_2}-n_{N_2})$ in the initial mixture ?

Answer 1

8

Answer 2

Let $V = 12.315$ litre be the volume of each bulb.

Initial state: Bulb A has $NH_3$, $CO_2$, $N_2$ at $T_A = 27^\circ C = 300$ K and total pressure $P_A = 60$ atm. Bulb B is at $T_B = -73^\circ C = 200$ K, Bulb C is at $T_C = -173^\circ C = 100$ K. Bulbs B and C are initially empty.

Let $n_{NH_3}$, $n_{CO_2}$, $n_{N_2}$ be the initial moles of $NH_3$, $CO_2$, and $N_2$ in bulb A. Initial total moles in bulb A: $n_A = n_{NH_3} + n_{CO_2} + n_{N_2}$. Using the ideal gas law for bulb A initially: $P_A V = n_A R T_A$. $60 \times 12.315 = n_A R \times 300$. $n_A R = \frac{60 \times 12.315}{300} = \frac{12.315}{5} = 2.463$. So, $n_{NH_3} R + n_{CO_2} R + n_{N_2} R = 2.463$.

Case 1: Stop-cock (i) is opened. Bulbs A (300 K) and B (200 K) are connected. The total volume is $V_A + V_B = 2V$. Liquefaction points: $NH_3$ at 253 K, $CO_2$ at 193 K, $N_2$ at 93 K. At $T_B = 200$ K: $NH_3$ (liquefies at 253 K): $200 < 253$, so $NH_3$ can condense. Vapour pressure of $NH_3(l)$ at 200 K is 1 atm. $CO_2$ (liquefies at 193 K): $200 > 193$, so $CO_2$ remains as gas. Vapour pressure of $CO_2(l)$ is negligible (meaning if liquid is present, gas pressure is negligible), but at 200 K it's a gas. $N_2$ (liquefies at 93 K): $200 > 93$, so $N_2$ remains as gas.

After opening stop-cock (i), the pressure in containers becomes $P_1 = 17$ atm. In bulb A (300 K), all gases are in gaseous state. In bulb B (200 K), $CO_2$ and $N_2$ are in gaseous state. $NH_3$ can be in gaseous and liquid state. Let $P'_{NH_3}$, $P'_{CO_2}$, $P'_{N_2}$ be the partial pressures of the gases in the gaseous phase. The total pressure is $P_1 = P'_{NH_3} + P'_{CO_2} + P'_{N_2} = 17$. If $NH_3$ condenses in bulb B, its partial pressure in the gas phase is equal to the vapour pressure at 200 K, which is 1 atm. So $P'_{NH_3} = 1$ atm. Then $P'_{CO_2} + P'_{N_2} = 17 - 1 = 16$ atm.

Assuming $P'_{NH_3} = 1$ atm. Total moles of $NH_3$: $n_{NH_3} = n'_{NH_3, A} + n'_{NH_3, B} + n'_{NH_3, liquid}$. $n'_{NH_3, A} = \frac{P'_{NH_3} V}{R T_A} = \frac{1 \times V}{R \times 300}$. $n'_{NH_3, B} = \frac{P'_{NH_3} V}{R T_B} = \frac{1 \times V}{R \times 200}$. Moles of $NH_3$ in gas phase after opening (i): $n'_{NH_3, gas} = n'_{NH_3, A} + n'_{NH_3, B} = \frac{V}{R} (\frac{1}{300} + \frac{1}{200}) = \frac{V}{R} \frac{5}{600} = \frac{V}{120R}$. Total initial moles of $NH_3$ is $n_{NH_3} = \frac{P_{NH_3, A} V}{R T_A}$. If $P_{NH_3, A} > \frac{T_A}{120} = \frac{300}{120} = 2.5$, then condensation occurs.

Total moles of $CO_2$: $n_{CO_2} = n'_{CO_2, A} + n'_{CO_2, B} = \frac{P'_{CO_2} V}{R T_A} + \frac{P'_{CO_2} V}{R T_B} = \frac{P'_{CO_2} V}{R} (\frac{1}{300} + \frac{1}{200}) = \frac{P'_{CO_2} V}{120R}$. Total moles of $N_2$: $n_{N_2} = n'_{N_2, A} + n'_{N_2, B} = \frac{P'_{N_2} V}{R T_A} + \frac{P'_{N_2} V}{R T_B} = \frac{P'_{N_2} V}{R} (\frac{1}{300} + \frac{1}{200}) = \frac{P'_{N_2} V}{120R}$.

From the initial state, $n_{CO_2} = \frac{P_{CO_2, A} V}{R T_A} = \frac{P_{CO_2, A} V}{300R}$. So, $\frac{P_{CO_2, A} V}{300R} = \frac{P'_{CO_2} V}{120R} \implies P_{CO_2, A} = \frac{300}{120} P'_{CO_2} = 2.5 P'_{CO_2}$. Similarly, $P_{N_2, A} = 2.5 P'_{N_2}$. $P_{CO_2, A} + P_{N_2, A} = 2.5 (P'_{CO_2} + P'_{N_2}) = 2.5 \times 16 = 40$ atm. Initial partial pressure of $NH_3$ in A: $P_{NH_3, A} = 60 - (P_{CO_2, A} + P_{N_2, A}) = 60 - 40 = 20$ atm. Since $P_{NH_3, A} = 20 > 2.5$, condensation of $NH_3$ occurs in bulb B, so our assumption $P'_{NH_3} = 1$ atm is correct.

Initial moles of each gas: $n_{NH_3} = \frac{P_{NH_3, A} V}{R T_A} = \frac{20 V}{300 R} = \frac{V}{15 R}$. $n_{CO_2} = \frac{P_{CO_2, A} V}{R T_A} = \frac{2.5 P'_{CO_2} V}{300 R} = \frac{P'_{CO_2} V}{120 R}$. $n_{N_2} = \frac{P_{N_2, A} V}{R T_A} = \frac{2.5 P'_{N_2} V}{300 R} = \frac{P'_{N_2} V}{120 R}$. $n_{CO_2} + n_{N_2} = \frac{(P'_{CO_2} + P'_{N_2}) V}{120 R} = \frac{16 V}{120 R} = \frac{2 V}{15 R}$.

Check total initial moles: $n_A = n_{NH_3} + n_{CO_2} + n_{N_2} = \frac{V}{15 R} + \frac{2 V}{15 R} = \frac{3 V}{15 R} = \frac{V}{5 R}$. From $60 V = n_A R T_A = n_A R \times 300$, $n_A R = \frac{60 V}{300} = \frac{V}{5}$. So $n_A = \frac{V}{5R}$. This matches.

Case 2: Both stop-cocks (i) and (ii) are opened. Bulbs A (300 K), B (200 K), and C (100 K) are connected. The total volume is $V_A + V_B + V_C = 3V$. The pressure decreases to $P_2 = 4$ atm. At $T_C = 100$ K: $NH_3$ (liquefies at 253 K): $100 < 253$, can condense. Vapour pressure of $NH_3(l)$ at 100 K is negligible. $CO_2$ (liquefies at 193 K): $100 < 193$, can condense. Vapour pressure of $CO_2(l)$ is negligible. $N_2$ (liquefies at 93 K): $100 > 93$, remains as gas.

In the final state, the partial pressures of $NH_3$ and $CO_2$ in the gas phase are negligible (close to 0) if liquid is present. Since the total pressure is 4 atm, and the partial pressures of $NH_3$ and $CO_2$ are negligible, the partial pressure of $N_2$ must be approximately 4 atm. Let $P''_{NH_3}$, $P''_{CO_2}$, $P''_{N_2}$ be the partial pressures in the gas phase. $P''_{NH_3} + P''_{CO_2} + P''_{N_2} = 4$. Given that vapour pressure of $NH_3(l)$ at 100 K is negligible, $P''_{NH_3} \approx 0$. Given that vapour pressure of $CO_2(l)$ is negligible, $P''_{CO_2} \approx 0$. So, $P''_{N_2} \approx 4$ atm.

Let's check if $N_2$ remains as gas. The partial pressure of $N_2$ is 4 atm. The temperature is 100 K. The liquefaction point is 93 K. Since the lowest temperature is 100 K > 93 K, $N_2$ remains as gas in all bulbs.

Total moles of $N_2$: $n_{N_2} = n''_{N_2, A} + n''_{N_2, B} + n''_{N_2, C}$. $n''_{N_2, A} = \frac{P''_{N_2} V}{R T_A} = \frac{4 V}{R \times 300}$. $n''_{N_2, B} = \frac{P''_{N_2} V}{R T_B} = \frac{4 V}{R \times 200}$. $n''_{N_2, C} = \frac{P''_{N_2} V}{R T_C} = \frac{4 V}{R \times 100}$. $n_{N_2} = \frac{4 V}{R} (\frac{1}{300} + \frac{1}{200} + \frac{1}{100}) = \frac{4 V}{R} (\frac{2 + 3 + 6}{600}) = \frac{4 V}{R} \frac{11}{600} = \frac{44 V}{600 R} = \frac{11 V}{150 R}$.

We had $n_{N_2} = \frac{2.5 P'_{N_2} V}{300 R} = \frac{P'_{N_2} V}{120 R}$. So, $\frac{P'_{N_2} V}{120 R} = \frac{11 V}{150 R} \implies P'_{N_2} = \frac{11 \times 120}{150} = \frac{11 \times 12}{15} = \frac{11 \times 4}{5} = 8.8$ atm. Then $P'_{CO_2} = 16 - P'_{N_2} = 16 - 8.8 = 7.2$ atm.

Now we can calculate the initial moles. $n_{NH_3} = \frac{V}{15 R}$. $n_{CO_2} = \frac{P'_{CO_2} V}{120 R} = \frac{7.2 V}{120 R} = \frac{72 V}{1200 R} = \frac{6 V}{100 R} = \frac{3 V}{50 R}$. $n_{N_2} = \frac{P'_{N_2} V}{120 R} = \frac{8.8 V}{120 R} = \frac{88 V}{1200 R} = \frac{22 V}{300 R} = \frac{11 V}{150 R}$.

We need to find the value of $(n_{NH_3} + n_{CO_2} - n_{N_2})$. $(n_{NH_3} + n_{CO_2} - n_{N_2}) = \frac{V}{15 R} + \frac{3 V}{50 R} - \frac{11 V}{150 R}$. Find a common denominator, which is 150. $= \frac{10 V}{150 R} + \frac{9 V}{150 R} - \frac{11 V}{150 R} = \frac{(10 + 9 - 11) V}{150 R} = \frac{8 V}{150 R} = \frac{4 V}{75 R}$.

We know $n_A R = 2.463$, so $R = \frac{2.463}{n_A}$. Also $n_A = \frac{V}{5R}$. $R = \frac{2.463}{(\frac{V}{5R})} = \frac{2.463 \times 5 R}{V}$. $1 = \frac{2.463 \times 5}{V} \implies V = 12.315$. This confirms the value of $V$. $R = \frac{2.463}{n_A}$. We also have $n_A = \frac{V}{5R}$.

Let's express the result in terms of $V/R$. $\frac{V}{R} = \frac{12.315}{R}$. Using $R=0.0821$, $V/R = 12.315/0.0821 \approx 150$. Using $R=0.082$, $V/R = 12.315/0.082 \approx 150.18$. Let's use $V/R = 150$ as a round number implied by the problem. $n_{NH_3} = \frac{150}{15} = 10$. $n_{CO_2} = \frac{3 \times 150}{50} = 3 \times 3 = 9$. $n_{N_2} = \frac{11 \times 150}{150} = 11$. Check total moles: $10 + 9 + 11 = 30$. $n_A = \frac{V}{5R} = \frac{150}{5} = 30$. This matches.

The value of $(n_{NH_3} + n_{CO_2} - n_{N_2}) = 10 + 9 - 11 = 19 - 11 = 8$.

Final check of assumptions: In Case 1, $P'_{NH_3} = 1$. $n'_{NH_3, gas} = \frac{1 \times V}{120R} = \frac{150}{120} = 1.25$. Initial moles $n_{NH_3} = 10$. Since $10 > 1.25$, condensation occurs in B. In Case 2, $P''_{NH_3} = 0$. $n''_{NH_3, gas} = \frac{0 \times V}{R} (\frac{1}{300} + \frac{1}{200} + \frac{1}{100}) = 0$. This implies all $NH_3$ is condensed. Total moles of $NH_3$ is 10. This is possible if the partial pressure is negligible. In Case 2, $P''_{CO_2} = 0$. $n''_{CO_2, gas} = 0$. This implies all $CO_2$ is condensed. Total moles of $CO_2$ is 9. This is possible if the partial pressure is negligible. In Case 2, $P''_{N_2} = 4$. $n''_{N_2, gas} = \frac{4 V}{150 R} = \frac{4 \times 150}{150} = 4$. Total moles of $N_2$ is 11. So, $11 - 4 = 7$ moles of $N_2$ are not in the gas phase. This is a contradiction, as $N_2$ should remain as gas at 100 K.

Let's re-evaluate the partial pressure of $N_2$ in Case 2. Total moles of $N_2$ is $n_{N_2} = 11$. After opening both stop-cocks, $N_2$ is distributed in A, B, and C as gas. $n_{N_2} = n''_{N_2, A} + n''_{N_2, B} + n''_{N_2, C} = \frac{P''_{N_2} V}{R T_A} + \frac{P''_{N_2} V}{R T_B} + \frac{P''_{N_2} V}{R T_C}$. $11 = \frac{P''_{N_2} V}{R} (\frac{1}{300} + \frac{1}{200} + \frac{1}{100}) = \frac{P''_{N_2} V}{R} \frac{11}{600}$. $11 = P''_{N_2} \frac{V}{R} \frac{11}{600} = P''_{N_2} \times 150 \times \frac{11}{600} = P''_{N_2} \times \frac{11}{4}$. $P''_{N_2} = 4$ atm. This is consistent with the total pressure being 4 atm if $NH_3$ and $CO_2$ partial pressures are negligible.

Let's re-evaluate the partial pressure of $NH_3$ in Case 2. Total moles of $NH_3$ is $n_{NH_3} = 10$. Partial pressure of $NH_3$ in gas phase is $P''_{NH_3}$. Moles of $NH_3$ in gas phase: $n''_{NH_3, gas} = \frac{P''_{NH_3} V}{R} (\frac{1}{300} + \frac{1}{200} + \frac{1}{100}) = \frac{P''_{NH_3} V}{R} \frac{11}{600}$. $n''_{NH_3, gas} = P''_{NH_3} \times 150 \times \frac{11}{600} = P''_{NH_3} \times \frac{11}{4}$. Given vapour pressure of $NH_3(l)$ at 100 K is negligible, $P''_{NH_3}$ should be negligible if liquid is present. If $P''_{NH_3} \approx 0$, then $n''_{NH_3, gas} \approx 0$. This means almost all $NH_3$ is condensed. This is consistent with $n_{NH_3} = 10$.

Let's re-evaluate the partial pressure of $CO_2$ in Case 2. Total moles of $CO_2$ is $n_{CO_2} = 9$. Partial pressure of $CO_2$ in gas phase is $P''_{CO_2}$. Moles of $CO_2$ in gas phase: $n''_{CO_2, gas} = \frac{P''_{CO_2} V}{R} (\frac{1}{300} + \frac{1}{200} + \frac{1}{100}) = \frac{P''_{CO_2} V}{R} \frac{11}{600} = P''_{CO_2} \times \frac{11}{4}$. Given vapour pressure of $CO_2(l)$ is negligible, $P''_{CO_2}$ should be negligible if liquid is present. If $P''_{CO_2} \approx 0$, then $n''_{CO_2, gas} \approx 0$. This means almost all $CO_2$ is condensed. This is consistent with $n_{CO_2} = 9$.

So, the assumption that partial pressures of $NH_3$ and $CO_2$ are negligible in Case 2 is consistent with the calculated initial moles.

The value of $(n_{NH_3} + n_{CO_2} - n_{N_2}) = 10 + 9 - 11 = 8$.

The final answer is $\boxed{8}$.

Explanation:

1. Calculate the initial total moles of gas in bulb A using the ideal gas law.
2. Analyze the state of each gas when stop-cock (i) is opened, considering liquefaction points and vapour pressures at the given temperatures. Determine which gases remain as gas and which may condense.
3. Use the given pressure after opening stop-cock (i) and the conditions in bulbs A and B to set up equations relating the initial moles and partial pressures. Assume condensation of $NH_3$ occurs and verify this assumption.
4. Analyze the state of each gas when both stop-cocks are opened, considering liquefaction points and vapour pressures at the given temperatures. Determine which gases remain as gas and which condense.
5. Use the given pressure after opening both stop-cocks and the conditions in bulbs A, B, and C to set up equations relating the initial moles and partial pressures. Assume condensation of $NH_3$ and $CO_2$ occurs and verify this assumption.
6. Solve the system of equations to find the initial moles of each gas.
7. Calculate the required value $(n_{NH_3} + n_{CO_2} - n_{N_2})$.

The final answer is $\boxed{8}$.