Question

Two vessels A and B, thermally insulated, contain an ideal monoatomic gas. A small tube fitted with a value connects these vessels. Initially the vessel A has 2 litres of gas at 300K and 2 xx 10^(5)Nm^(-2) pressure while vessel B has 4 litres of gas at 350 K and 4 xx 10^(5)Nm^(-2) pressure. The value is now opened and the system reaches equilibrium in pressure and temperature. Calculate the new pressure and temperature . (R = (25)/(3)J//mol-K)

Answer 1

Answer:

Final temperature of the mixture is

$T = 338.7 K$

final pressure is given as

$P = 3.33 \times 10^5 Pa$

Explanation:

As we know that the total internal energy of the system will remain same

so energy given by gas A = energy gain by gas B

so here we have

$PV = nRT$

$(2\times 10^5)(2\times 10^{-3}) = n_a (8.31)(300)$

$n_a = 0.16 moles$

Similarly for gas B

$PV = nRT$

$(4\times 10^5)(4\times 10^{-3}) = n_a (8.31)(350)$

$n_a = 0.55 moles$

Now by energy balance

$n_a R (T - 300) = n_b R (350 - T)$

$0.16 T - 48 = 192.5 - 0.55T$

$0.71 T = 240.5$

$T = 338.7 K$

now final pressure is given as

$\frac{P(V_1 + V_2)}{RT} = \frac{P_1V_1}{RT_1} + \frac{P_2V_2}{RT_2}$

here we have

$\frac{P(2 + 4)}{R(338.7)} = \frac{(2\times 10^5)(2)}{R(300)} + \frac{(4\times 10^5)(4)}{R(350)}$

$P = 3.33 \times 10^5 Pa$

#Learn

Topic : Ideal gas equation

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