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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