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Two glass bulbs A and B of volume V, 2V respectively are connected by a narrow capillary tube. The bulbs contain gas at temperature T and pressure P. Now the temperature of bulb A is doubled and that of B is tripled. The final number of moles of gas in bulb 'A' is

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$\frac{3}{2} \frac{P V}{R T}$

$\frac{5}{7} \frac{P V}{R T}$

$\frac{12}{7} \frac{P V}{R T}$

$\frac{9}{7} \frac{P V}{R T}$

answer is D.

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Detailed Solution

Initial no of moles

$n_{1} = \frac{P V}{R T} n_{2} = \frac{P (2 V)}{R T}$

Final no of moles.

$n_{1}^{1} = \frac{P^{1} V}{R (2 T)} n_{2}^{1} = \frac{P^{1} (2 V)}{R (3 T)}$

$n_{1} + n_{2} = n_{1}^{1} + n_{2}^{1}$

$\frac{3 P V}{R T} = \frac{P^{1} V}{R T} (\frac{1}{2} + \frac{2}{3})$

$3 P = P^{1} (\frac{3 + 4}{6})$

$P^{1} = \frac{18}{7} P_{1}^{1} = \frac{18 P}{7} \frac{V}{R 2 T} = \frac{9 P V}{7 R T}$

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