Two spherical vessel of equal volume, are connected by a n arrow tube. The apparatus contains an ideal gas at one atmosphere and 300K. Now if one vessel is immersed in a bath of constant temperature 600K and the other in a bath of constant temperature 300K. Then the common pressure will be

1 a t m

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\frac{4}{5} a t m

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\frac{4}{3} a t m

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\frac{3}{4} a t m

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Solution

The correct option is C $\frac{4}{3} a t m$
$μ = m u_{1} + m u_{2}$
$\frac{P (2 V)}{R T_{1}} = \frac{P^{'} V}{R T_{1}} + \frac{P^{'} V}{R T_{2}} \Rightarrow \frac{2 P}{R T_{1}} = \frac{P^{'}}{R} [\frac{T_{2} + T_{1}}{T_{1} T_{2}}]$
$P^{'} = \frac{2 P T_{2}}{(T_{1} + T_{2})} = \frac{2 \times 1 \times 600}{(300 + 600)} = \frac{4}{3} a t m$ .
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Two spherical vessel of equal volume, are connected by a n arrow tube. The apparatus contains an ideal gas at one atmosphere and 300K. Now if one vessel is immersed in a bath of constant temperature 600K and the other in a bath of constant temperature 300K. Then the common pressure will be

The correct option is C 43 atmμ=mu1+mu2 P(2V)RT1=P′VRT1+P′VRT2⇒2PRT1=P′R[T2+T1T1T2] P′=2PT2(T1+T2)=2×1×600(300+600)=43atm. .