# An #8 L# container holds #8 # mol and #9 # mol of gasses A and B, respectively. Every two molecules of gas B bind to one molecule of gas A and the reaction raises the temperature from #370^oK# to #425 ^oK#. How much does the pressure change by?

30.377atm at 370K

34.8925atm at 425K

So this is chemistry stoichiometry

So the reagent in excess is A and some amount of it will remain till the end

So the amount of AB2 is equal to the amount of A used

So A used is 4.5mol

But left out is 8 - 4.5 = 3.5mol

So mol of A + mol of AB2 =

4.5mol + 3.5mol = 8mol

Now calculate the volume of gas at 370K and 425K

using the formula

Where P is pressure in atm n is moles R is the universal gas constant which is equal to 0.0821L T is temperature in kelvin V is volume

Plug in the variables

Now calculate the pressure at different temperature

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To solve this problem, we can use the ideal gas law and the concept of stoichiometry. The reaction between gases A and B results in the formation of a new compound, which may affect the pressure inside the container. Since two molecules of gas B bind to one molecule of gas A, the total decrease in the number of gas molecules will affect the pressure. We can use the combined gas law to find the initial and final pressures, then calculate the pressure change.

Using the ideal gas law:

Initial pressure, ( P_1 = \frac{n_1RT_1}{V} )

Final pressure, ( P_2 = \frac{n_2RT_2}{V} )

Where:

- ( n_1 ) and ( n_2 ) are the initial and final number of moles of gas (A and B together).
- ( R ) is the ideal gas constant.
- ( T_1 ) and ( T_2 ) are the initial and final temperatures.
- ( V ) is the volume.

The ratio of gas B to gas A is 9:8. So, if 9 mol of gas B reacts, 4.5 mol of gas A will also react.

( n_1 = 8 , \text{mol} + 9 , \text{mol} = 17 , \text{mol} )

( n_2 = 8 , \text{mol} + (9 - 4.5) , \text{mol} = 13.5 , \text{mol} )

Plugging in the values:

( P_1 = \frac{17 , \text{mol} \times 0.0821 , \text{atm} \cdot \text{L} / \text{mol} \cdot \text{K} \times 370 , \text{K}}{8 , \text{L}} \approx 52.3 , \text{atm} )

( P_2 = \frac{13.5 , \text{mol} \times 0.0821 , \text{atm} \cdot \text{L} / \text{mol} \cdot \text{K} \times 425 , \text{K}}{8 , \text{L}} \approx 71.6 , \text{atm} )

Now, we find the change in pressure:

( \Delta P = P_2 - P_1 = 71.6 , \text{atm} - 52.3 , \text{atm} \approx 19.3 , \text{atm} )

Therefore, the pressure increases by approximately 19.3 atm.

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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.

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