A #3 L# container holds #5 # mol and #5 # 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 #340^oK# to #320^oK#. How much does the pressure change?
We have to take two things into account:
the change in moles and the change in temperature.
Note: Of course you could have used the general gas-formula
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To find the change in pressure, you can use the ideal gas law, which states:
PV = nRT
Where: P = pressure V = volume n = number of moles R = ideal gas constant T = temperature in Kelvin
Given: Initial temperature (T1) = 340 K Final temperature (T2) = 320 K Initial number of moles of gas A (nA) = 5 mol Initial number of moles of gas B (nB) = 5 mol Initial total number of moles (n_total) = nA + nB Volume (V) = 3 L
To find the initial pressure (P1), you can use the initial conditions:
P1 = (n_total * R * T1) / V
To find the final pressure (P2), you need to use the final conditions:
P2 = (n_total * R * T2) / V
Then, the change in pressure (ΔP) can be calculated as:
ΔP = P2 - P1
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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.
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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