# A #6 L# container holds #4 # mol and #6 # mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from #480^oK# to #270^oK#. How much does the pressure change?

Final Pressure :

Pressure Change :

The fact that the pressure change is negative means that the pressure is dropping.

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To find the change in pressure, we can use the ideal gas law, which states:

[ PV = nRT ]

Where:

- ( P ) is the pressure,
- ( V ) is the volume,
- ( n ) is the number of moles,
- ( R ) is the gas constant,
- ( T ) is the temperature.

Since the volume, number of moles, and gas constant remain constant in this case, we can use the relationship between pressure and temperature directly:

[ \frac{P_2}{P_1} = \frac{T_2}{T_1} ]

Given that the initial temperature ( T_1 = 480 , \text{K} ) and the final temperature ( T_2 = 270 , \text{K} ), we can calculate the change in pressure using:

[ \frac{P_2}{P_1} = \frac{270 , \text{K}}{480 , \text{K}} ]

[ P_2 = P_1 \times \frac{270 , \text{K}}{480 , \text{K}} ]

Now, we need to find the initial pressure (( P_1 )). Since we have 4 moles of gas A and 6 moles of gas B in a 6 L container, the total number of moles is 10. Using the ideal gas law with the initial temperature (( T_1 = 480 , \text{K} )) and the number of moles (( n = 10 )):

[ P_1 = \frac{nRT}{V} = \frac{(10 , \text{mol}) \times (8.314 , \text{J/mol·K}) \times (480 , \text{K})}{6 , \text{L}} ]

Now, we can calculate ( P_2 ) using the formula we derived earlier:

[ P_2 = P_1 \times \frac{270 , \text{K}}{480 , \text{K}} ]

[ P_2 = \left( \frac{(10 , \text{mol}) \times (8.314 , \text{J/mol·K}) \times (480 , \text{K})}{6 , \text{L}} \right) \times \frac{270 , \text{K}}{480 , \text{K}} ]

After calculating ( P_1 ) and ( P_2 ), we can find the change in pressure (( \Delta P )):

[ \Delta P = P_2 - P_1 ]

Substitute the values to calculate the change in pressure.

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According to the ideal gas law, ( PV = nRT ), where ( P ) is pressure, ( V ) is volume, ( n ) is the number of moles, ( R ) is the gas constant, and ( T ) is temperature.

Since the volume and the number of moles of gas A and B remain constant, we can use the formula ( P_1/T_1 = P_2/T_2 ) to find the pressure change, where ( P_1 ) and ( T_1 ) are the initial pressure and temperature, and ( P_2 ) and ( T_2 ) are the final pressure and temperature, respectively.

Given that the initial temperature ( T_1 = 480 ) K and the final temperature ( T_2 = 270 ) K, we can use the formula to calculate the pressure change.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- A #6 L# container holds #4 # mol and #6 # mol of gasses A and B, respectively. Groups of three of molecules of gas B bind to two molecules of gas A and the reaction changes the temperature from #480^oK# to #270^oK#. How much does the pressure change?
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