A rigid vessel of volume 0.50 m3 containing H2 at 20.5oC and a pressure of 611 x 103 Pa is connected to a second rigid vessel of volume 0.75 m3 containing Ar at 31.2oC at a pressure of 433 x103 Pa. A valve separating the two vessels is opened and both are cooled to a temperature of 14.5oC. What is the fina1 pressure in the vessels?

nT=n1+n2P×1.25R×287.5=611×103×0.5293.5R+433×103×0.75204.5RP=4.84×105Pa

A rigid vessel of volume 0.50 m3 containing H2 at 20.5oC and a pressure of 611 x 103 Pa is connected to a second rigid vessel of volume 0.75 m3 containing Ar at 31.2oC at a pressure of 433 x103 Pa. A valve separating the two vessels is opened and both are cooled to a temperature of 14.5oC. What is the fina1 pressure in the vessels?

nT=n1+n2P×1.25R×287.5=611×103×0.5293.5R+433×103×0.75204.5RP=4.84×105Pa

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A rigid vessel of volume 0.50 m³ containing H₂ at 20.5^oC and a pressure of 611 x 10³ Pa is connected to a second rigid vessel of volume 0.75 m³ containing Ar at 31.2^oC at a pressure of 433 x10³ Pa. A valve separating the two vessels is opened and both are cooled to a temperature of 14.5^oC. What is the fina1 pressure in the vessels?

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2 x 10⁵

3.22 x 10⁵Pa

4840 Pa

4.84 x 10⁵Pa

answer is D.

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Detailed Solution

$\begin{array}{l} n_{T} = n_{1} + n_{2} \\ \frac{P \times 1 .25}{R \times 287 .5} = \frac{611 \times 10^{3} \times 0 .5}{293 .5 R} + \frac{433 \times 10^{3} \times 0 .75}{204 .5 R} \\ P = 4 .84 \times 10^{5} Pa \end{array}$

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