Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure Pi and temperature Ti are connected through a narrow tube of negligible volume as shown in the figure below. The pressure of one of the bulb is raised to T2. The final pressure is: (JEE Mains-2016)

 

Initially both bulbs are under similar condition hence number of moles of gas in each container

Total number of moles in both bulbs;

When temperature of one of the bulb is increases from T₁ toT₂ then final pressure P_f  and number of moles.  

A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does not allow the gas to leak across (Figure 2), the volume (in m3) of the compartment A after the system attains equilibrium is--- (IIT-Ad-2018)

Total volume of compartment A and B

 V_A+V_B = 4m³…………………………………….................................(1)

According to question partition do not allow gas leak between compartment (A) and compartment (B) so number of moles after and before inserting new partition which can slide.

So number of moles in compartment (A) and compartment (B)

n_A=PxV_A/RT=1m³x5 bar/400 K  ……………………………………… (2)

n_B=PxV_B/RT=3m³x1 bar/300 K  ………………………………………. (3)

Equation (2)/ (3)

Get V_A/V_B=5/4   Or  V_A=5V_B/4 ………………………………………..(4)

Put the value of V_A   from equation (4) to in equation (1) and find the value of V_A  after placing frictionless slider.

5V_B/4 +V_B = 4m³        => V_B = 16/9m³       and  V_A = 20/9m³   =2.22  m³  or L

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Initially when tube is horizontal then pressure (P) =76 cm Hg and volume (V) =45 cm of tube.

Finally when tube is made vertical, as temperature and pressure are constant, then by gas law

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