Avessel (see the figure) comprises of into two chambers
x and y with an adiabatic wall. the adiabatic wall can
move freely. the chambers x and y contain an ideal gas
under the condition that pressure p, volume v, and
temperature t are all same, respectively. the system
gains an equilibrium fate after heating when the
temperature of chamber x becomes 3t. while the
temperature of chamber y is always constant at t. the
gas pressure of chamber y at the equilibrium after the
heating is

The gas pressure of chamber Y at the equilibrium after the

heating is 2P.

This is defined as the equation of state of a hypothetical ideal gas and it is depicted by the following below:

PV = nRT

We can infer that the number of moles(n) and R (ideal gas constant) are constant

PV / T (Initial) = PV / T (Final)

We were told that Chamber X starts at pressure P, volume V, and temperature T.  At equilibrium, the pressure is Px, the volume is Vx, and temperature 3T.

PV / T = P V / 3T

We were also told that Chamber Y starts at pressure P, volume V, and temperature T.  At equilibrium, the pressure is Py, the volume is Vy, and temperature T.

PV / T = Py Vy / T

Substitute into the equation:

Px Vx / 3T = Py Vy / T

Px Vx / 3 = Py Vy

At equilibrium, the chambers are : Px = Py:

Vx / 3 = Vy

Vx = 3 Vy

The total volume is the same as before, so:

Vx + Vy = 2V

Substitute into the equation

(3 Vy) + Vy = 2V

4 Vy = 2V

Vy = V / 2

Substitute into our equation for chamber Y:

PV / T = Py (V/2) / T

PV = Py (V/2)

Py = 2P

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(C) 2P

Explanation:

Ideal gas law states:

PV = nRT

n (the number of moles) and R (ideal gas constant) are constant, so we can say:

(PV / T) before = (PV / T) after

Chamber X starts at pressure P, volume V, and temperature T.  At equilibrium, the pressure is Px, the volume is Vx, and temperature 3T.

PV / T = Px Vx / 3T

Chamber Y starts at pressure P, volume V, and temperature T.  At equilibrium, the pressure is Py, the volume is Vy, and temperature T.

PV / T = Py Vy / T

Substituting and simplifying:

Px Vx / 3T = Py Vy / T

Px Vx / 3 = Py Vy

Since the chambers are at equilibrium, Px = Py:

Vx / 3 = Vy

Vx = 3 Vy

The total volume is the same as before, so:

Vx + Vy = 2V

Substituting:

(3 Vy) + Vy = 2V

4 Vy = 2V

Vy = V / 2

Now if we substitute into our equation for chamber Y:

PV / T = Py (V/2) / T

PV = Py (V/2)

Py = 2P

The pressure in chamber Y (and chamber X) doubles at equilibrium.