Question 964: Thermodynamics & KTG

Question

A container has two chambers of volumes $V_1=2$ litres and $V_2=3$ litres separated by a partition made of a thermal insulator. The chambers contain $n_1=5$ and $n_2=4$ moles of ideal gas at pressures $p_1=1 \mathrm{~atm}$ and $p_2=2 \mathrm{~atm}$, respectively. When the partition is removed, the mixture attains an equilibrium pressure of

Accepted Answer

To find the equilibrium pressure of the system after the partition is removed, we can apply the principle of conservation of moles (or the ideal gas law in combined volumes).
Initially, we have:
Chamber 1: $ P_1 = 1 \, 	ext{atm}, \, V_1 = 2 \, 	ext{L} $
Chamber 2: $ P_2 = 2 \, 	ext{atm}, \, V_2 = 3 \, 	ext{L} $
The total pressure after the partition is removed and the gases mix can be calculated using:
$  P_{	ext{final}} = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2}  $
Substituting the given values:
$  P_{	ext{final}} = \frac{1 	imes 2 + 2 	imes 3}{2 + 3} = \frac{2 + 6}{5} = \frac{8}{5}  $
This gives us:
$  P_{	ext{final}} = 1.6 \, 	ext{atm}  $
Therefore, the equilibrium pressure when the gases mix is $ 1.6 \, 	ext{atm} $.

Answer

1.4 atm

Answer

1.8 atm

Answer

1.3 atm

Answer

1.6 atm

Thermodynamics & KTG

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