Thermodynamics

JEE Main level, single correct, medium

59 of 190

Question 59 of 190

Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure 9090 kPa and temperature 400400 K. Keeping the temperature of one vessel constant at 400400 K the second vessel temperature is raised to 500500 K. The final pressure in the vessels is _______ kPa.

Show answer and solution

Answer: (A) 100100

Let the volume of each vessel be VV.

Initially, both vessels are at pressure P1=90P_1 = 90 kPa and temperature T1=400T_1 = 400 K.

Total number of moles initially is given by:

ni=P1VRT1+P1VRT1=2P1VRT1n_i = \dfrac{P_1 V}{R T_1} + \dfrac{P_1 V}{R T_1} = \dfrac{2 P_1 V}{R T_1}

Finally, the temperature of the first vessel is T1=400T_1 = 400 K and the second vessel is T2=500T_2 = 500 K. Let the final pressure be PfP_f.

Total number of moles finally is given by:

nf=PfVRT1+PfVRT2n_f = \dfrac{P_f V}{R T_1} + \dfrac{P_f V}{R T_2}

Since the total number of moles remains constant, ni=nfn_i = n_f:

2P1VRT1=PfVRT1+PfVRT2\dfrac{2 P_1 V}{R T_1} = \dfrac{P_f V}{R T_1} + \dfrac{P_f V}{R T_2}

Substituting the given values:

2×90400=Pf400+Pf500\dfrac{2 \times 90}{400} = \dfrac{P_f}{400} + \dfrac{P_f}{500}

180400=Pf(1400+1500)\dfrac{180}{400} = P_f \left( \dfrac{1}{400} + \dfrac{1}{500} \right)

180400=Pf(5+42000)\dfrac{180}{400} = P_f \left( \dfrac{5 + 4}{2000} \right)

180400=Pf(92000)\dfrac{180}{400} = P_f \left( \dfrac{9}{2000} \right)

Pf=180×2000400×9P_f = \dfrac{180 \times 2000}{400 \times 9}

Pf=20×5=100P_f = 20 \times 5 = 100 kPa

Answer: 100100

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.

More Physics practice

JEE Main Physics practice question on Thermodynamics (Thermodynamic Processes), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.