NVAEasyJEE 2023Stoichiometry & Calculations

JEE Chemistry 2023 Question with Solution

Three bulbs are filled with CH4_4, CO2_2, and Ne as shown in the picture. The bulbs are connected through pipes of zero volume. When the stopcocks are opened and the temperature is kept constant throughout, the pressure of the system is found to be _____ atm. (Nearest integer)

Three connected bulbs containing CH4, CO2, and Ne gases with pressures 2 atm, 4 atm, and 3 atm and volumes 2 L, 3 L, and 4 L respectively, joined by zero-volume pipes with stopcocks.

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: P1=2atm,V1=2LP_1 = 2 \, \text{atm}, V_1 = 2 \, \text{L} for CH4_4, P2=4atm,V2=3LP_2 = 4 \, \text{atm}, V_2 = 3 \, \text{L} for CO2_2, and P3=3atm,V3=4LP_3 = 3 \, \text{atm}, V_3 = 4 \, \text{L} for Ne. Temperature remains constant and the connecting pipes have zero volume.

Find: The final pressure of the system.

Using Dalton's law:

PfVf=P1V1+P2V2+P3V3P_f V_f = P_1 V_1 + P_2 V_2 + P_3 V_3

Total volume:

Vf=V1+V2+V3=2+3+4=9LV_f = V_1 + V_2 + V_3 = 2 + 3 + 4 = 9 \, \text{L}

Total pressure contribution:

PfVf=(2×2)+(4×3)+(3×4)=4+12+12=28atm.LP_f V_f = (2 \times 2) + (4 \times 3) + (3 \times 4) = 4 + 12 + 12 = 28 \, \text{atm.L}

Final pressure:

Pf=2893.11atmP_f = \frac{28}{9} \approx 3.11 \, \text{atm}

Thus, Pf3P_f \approx 3 atm.

Common mistakes

  • Adding the pressures directly as 2+4+32 + 4 + 3 is incorrect because the gases are initially in different volumes. First use PfVf=PiViP_f V_f = \sum P_i V_i, then divide by the total volume.

  • Ignoring the total final volume is wrong. After opening the stopcocks, all gases occupy the combined volume 2+3+4=9L2 + 3 + 4 = 9 \, \text{L}, not their original separate volumes.

  • Using the identity of gases to change the calculation is unnecessary here. At constant temperature, the final pressure depends on the pressure-volume contributions, not on whether the gas is CH4_4, CO2_2, or Ne.

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