Q131CP - Chemistry: Molecular Nature Of Matter And Change

Question

Question: On a humid day in New Orleans, the temperature is  , and the partial pressure of water vapor in the air is 31.0 torr. The 9000-ton air-conditioning system in the Louisiana Superdome maintains an inside air temperature of  also, but a partial pressure of water vapor of 10.0 torr. The volume of air in the dome is  , and the total pressure inside and outside the dome are both 1.0 atm.

a. What mass of water (in metric tons) must be removed every time the inside air is completely replaced with outside air? 

(Hint: How many moles of gas are in the dome? How many moles of water vapor? How many moles of dry air? How many moles of outside air must be added to the air in the dome to simulate the composition of outside air?)

b. Find the heat released when this mass of water condenses.

 

Step-by-Step Solution

Answer

Answer

 

1. The mass of water that must be removed every time the inside air is completely replaced with outside air in the dome can be 73.2ton.
2. The heat released when the given mass of water condenses is$9.07\times {10}^{7}J$  .

 

Step 1: Definition

The ideal gas does not affect by the change in the pressure or volume of the gas at any temperature. The equation for the ideal gas:

PV = nRT

The number of moles can be defined as the ratio of the mass of the atom/molecule and the molar mass of the atom/molecule.

 $\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{ }\mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{MolarMass}}$ 

Heat energy can be defined as the rise in temperature required to raise the heat of the specific mass of the matter.

 

The equation for the Heat energy change:

 $\mathbf{\Delta H}\mathbf{=}\mathbf{m}\mathbf{\times }\mathbf{S}\mathbf{\times }\mathbf{\Delta T}$

Here, S is the Specific heat of a solid, M is Mass and T is Temperature.

Subpart (a) The mass of water that must be removed.

As given,

Outside in New Orleans,

Temperature is  ${\mathbf{\text{22.0}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$.

The partial pressure of water vapor in the air is 31.0 torr. 

 

Inside the dome, a 9000-ton AC system

Temperature is ${\mathbf{\text{22.0}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$ .

The partial pressure of water vapor is 10.0 torr.

The volume of air in the dome is $2.4\times {10}^6{m}^2$ .

 

The total pressure inside and outside is 1.0 atm.

 

Step 1: Volume of water in the dome.

The temperature on a humid day  $={22}^{o}C=295k$

$$\begin{gathered} Volumeofthewater=2.4\times {10}^6{m}^3 \\ 1L={10}^3{m}^3 \\ Volumeofthewater=2.4\times {10}^6{m}^3\times \frac{1L}{{10}^{-3}{m}^3} \\ Volumeofthewater=2.4\times {10}^{9}L \end{gathered}$$ 

   

 

The volume of water in the dome  $=2.4\times {10}^{9}L$

 

Step 2: The pressure at a given temp.

Gas Constant,  $R=8.314J/k/mole$

 

Convert the torr into atmospheric pressure:

 

Pressure at temperature  31 torr

 $$\begin{gathered} 1 atm=760 torr \\ \mathrm{Pr}essure at temperature=31 torr\times \frac{1 atm}{760 torr} \\ \mathrm{Pr}essure at temperature=0.041 atm \end{gathered}$$

 

 

 

Step 3: Number of moles of dry air.

 $$\begin{gathered} PV=nRT \\ 0.041\times 2.4\times {10}^{9}L=n\times 8.314\times 295 \\ n=\frac{0.041\times 2.4\times {10}^{9}L}{8.314\times 295} \\ n=4.06\times {10}^6 \end{gathered}$$

Number of moles of dry air  $=4.06\times {10}^{6}mole$

 

Step 4: The mass of water that must be removed.

Molar mass of water molecule = 18g/mole

 

Mass of the water molecule at  :22⁰C

 $$\begin{gathered} \mathbf{Number} \mathbf{of} \mathbf{Moles}\mathbf{=}\frac{\mathbf{Mass}}{\mathbf{MolarMass}} \\ 4.06\times {10}^{6}mole=\frac{Mass}{18} \\ Mass=18\times 4.06\times {10}^6 \\ Mass=7.32\times {10}^{7}g \end{gathered}$$ 

Conversion of gram into ton:

 $$\begin{gathered} 1 kg=1000 g \\ 1 ton=1000 kg \\ Mass=7.32\times {10}^{7}g\times \frac{1 ton}{1000}\times \frac{1 kg}{1000 g} \\ Mass=73.2 ton \end{gathered}$$

 

 

The mass of water that must be removed in the dome can be 73.2ton.