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Q17P

Expert-verifiedFound in: Page 578

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: Container A in figure holds an ideal gas at a pressure of ${\mathbf{5}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{5}}}{\mathbf{Pa}}$**** and a temperature of 300 k ****It is connected by a thin tube (and a closed valve) to container B, with four times the volume of A. Container B holds the same ideal gas at a pressure of ${\mathbf{1}}{\mathbf{.}}{\mathbf{0}}{\mathbf{\times}}{{\mathbf{10}}}^{{\mathbf{5}}}{\mathbf{Pa}}$**** and a temperature of 400 k****. The valve is opened to allow the pressures to equalize, but the temperature of each container is maintained. What then is the pressure?**

**Answer**

Pressure of the container after the valve is opened is.$2.0\times {10}^{5}\text{Pa}$

** **

Write the total number of moles in terms of the volume of gas in container A using the gas law. Now, equate the pressures **after the valve is opened and find the number of moles of container B, after the valve is opened. Substitute this in the expression for total number of moles after the valve is opened, find the pressure in the containers.**

Formula is as follow:

${p}_{i}{v}_{i}=nR{T}_{i}$

Here, *p* is pressure, *v* is volume, *T* is temperature, *R* is universal gas constant and *n* is number of moles.

** **

The total number of molecules in both the containers is,

$n={n}_{A}+{n}_{B}$

According to the gas law,

$pV=nRT\phantom{\rule{0ex}{0ex}}\therefore n=\frac{{p}_{A}{V}_{A}}{R{T}_{A}}+\frac{{p}_{B}{V}_{B}}{R{T}_{B}}\phantom{\rule{0ex}{0ex}}\therefore n=\frac{5.0\times {10}^{5}\times {V}_{A}}{300\left(8.314\right)}+\frac{1.0\times {10}^{5}\times 4\times {V}_{A}}{400\left(8.314\right)}\phantom{\rule{0ex}{0ex}}\therefore n=320.744{V}_{A}$

For the valve is opened solve as:

role="math" localid="1661839503686" $p{\text{'}}_{A}=p{\text{'}}_{B}\phantom{\rule{0ex}{0ex}}\frac{n{\text{'}}_{A}R{T}_{A}}{{V}_{A}}=\frac{n{\text{'}}_{B}R{T}_{B}}{{V}_{B}}\phantom{\rule{0ex}{0ex}}\therefore n{\text{'}}_{B}=\frac{n{\text{'}}_{A}{T}_{A}{V}_{B}}{{T}_{B}{V}_{A}}\phantom{\rule{0ex}{0ex}}But{V}_{B}=4{V}_{A},\phantom{\rule{0ex}{0ex}}\therefore n{\text{'}}_{B}=\frac{4n{\text{'}}_{A}{T}_{A}}{{T}_{B}}$

The total no. of molecules in both the containers after valve is opened is,

$n=n{\text{'}}_{A}+n{\text{'}}_{B}\phantom{\rule{0ex}{0ex}}\frac{n{\text{'}}_{A}R{T}_{A}}{{V}_{A}}=\frac{n{\text{'}}_{B}R{T}_{B}}{{V}_{B}}\phantom{\rule{0ex}{0ex}}320.744{V}_{A}=n{\text{'}}_{A}+\frac{4n{\text{'}}_{A}{T}_{A}}{{T}_{B}}$

Substitute the values and solve:

$320.744{V}_{A}=\left(1+\frac{4{T}_{A}}{{T}_{B}}\right)n{\text{'}}_{A}\phantom{\rule{0ex}{0ex}}n{\text{'}}_{A}=\frac{320.744{V}_{A}}{\left(1+\frac{4{T}_{A}}{{T}_{B}}\right)}$

Pressure in the container A after valve is opened is,

$p{\text{'}}_{A}=\frac{n{\text{'}}_{A}R{T}_{A}}{{V}_{A}}\phantom{\rule{0ex}{0ex}}p{\text{'}}_{A}=\frac{\frac{320.744{V}_{A}}{\left(1+\frac{4{T}_{A}}{{T}_{B}}\right)}R{T}_{A}}{{V}_{A}}\phantom{\rule{0ex}{0ex}}p{\text{'}}_{A}=\frac{\frac{320.744{V}_{A}}{\left(1+\frac{4\left(300\right)}{400}\right)}\left(8.314\right)\left(300\right)}{{V}_{A}}\phantom{\rule{0ex}{0ex}}p{\text{'}}_{A}=p{\text{'}}_{B}=2.0\times {10}^{5}\text{Pa}$

Hence, pressure of the container after the valve is opened is.$2.0\times {10}^{5}\text{Pa}$

Therefore, the pressure of the mixture of the gas can be found from pressure, volume, and temperature of the individual gases.

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