JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 28)
A container is divided into two chambers by a partition. The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contain 4.0 moles of gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x $$\times$$ 10$$-$$1 atm. Value of x is ________.
إجابة
25
توضيح
By energy conservation
$${3 \over 2}{n_1}R{T_1} + {3 \over 2}{n_2}R{T_2} = {3 \over 2}({n_1} + {n_2})RT$$
Using PV = nRT
P1V1 + P2V2 = P(V1 + V2)
$$P = {{{P_1}{V_1} + {P_2}{V_2}} \over {{V_1} + {V_2}}} = {{2 \times 4.5 + 3 \times 5.5} \over {4.5 + 5.5}}$$
$$P = {{9 + 16.5} \over {10}} = {{25.5} \over {10}}$$
$$ \approx 25 \times {10^{ - 1}}$$ atm
$${3 \over 2}{n_1}R{T_1} + {3 \over 2}{n_2}R{T_2} = {3 \over 2}({n_1} + {n_2})RT$$
Using PV = nRT
P1V1 + P2V2 = P(V1 + V2)
$$P = {{{P_1}{V_1} + {P_2}{V_2}} \over {{V_1} + {V_2}}} = {{2 \times 4.5 + 3 \times 5.5} \over {4.5 + 5.5}}$$
$$P = {{9 + 16.5} \over {10}} = {{25.5} \over {10}}$$
$$ \approx 25 \times {10^{ - 1}}$$ atm