Q17P Question: Container A in figure ... [FREE SOLUTION]

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Chapter 19: Q17P (page 578)

Question: Container A in figure holds an ideal gas at a pressure of $5.0 脳 10^{5} Pa$ and a temperature of 300 kIt 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 $1.0 脳 10^{5} Pa$ 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?

Short Answer

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Answer

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

Step by step solution

Determining the concepts

Write thetotal 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:

$p_{i} v_{i} = n R T_{i}$

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

Determine the pressure of the container after the valve is opened

The total number of molecules in both the containers is,

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

According to the gas law,

$p V = n R T 鈭� n = \frac{p_{A} V_{A}}{R T_{A}} + \frac{p_{B} V_{B}}{R T_{B}} 鈭� n = \frac{5.0 脳 10^{5} 脳 V_{A}}{300 (8.314)} + \frac{1.0 脳 10^{5} 脳 4 脳 V_{A}}{400 (8.314)} 鈭� n = 320.744 V_{A}$

For the valve is opened solve as:

role="math" localid="1661839503686" $p'_{A} = p'_{B} \frac{n'_{A} R T_{A}}{V_{A}} = \frac{n'_{B} R T_{B}}{V_{B}} 鈭� n'_{B} = \frac{n'_{A} T_{A} V_{B}}{T_{B} V_{A}} B u t V_{B} = 4 V_{A}, 鈭� n'_{B} = \frac{4 n'_{A} T_{A}}{T_{B}}$

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

$n = n'_{A} + n'_{B} \frac{n'_{A} R T_{A}}{V_{A}} = \frac{n'_{B} R T_{B}}{V_{B}} 320.744 V_{A} = n'_{A} + \frac{4 n'_{A} T_{A}}{T_{B}}$

Substitute the values and solve:

$320.744 V_{A} = (1 + \frac{4 T_{A}}{T_{B}}) n'_{A} n'_{A} = \frac{320.744 V_{A}}{(1 + \frac{4 T_{A}}{T_{B}})}$

Pressure in the container A after valve is opened is,

$p'_{A} = \frac{n'_{A} R T_{A}}{V_{A}} p'_{A} = \frac{\frac{320.744 V_{A}}{(1 + \frac{4 T_{A}}{T_{B}})} R T_{A}}{V_{A}} p'_{A} = \frac{\frac{320.744 V_{A}}{(1 + \frac{4 (300)}{400})} (8.314) (300)}{V_{A}} p'_{A} = p'_{B} = 2.0 脳 10^{5} Pa$