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In summary, the conversation is about solving a problem using the ideal gas law equation PV=nRT, where "n" represents the number of moles. The participants discuss integrating both sides of the equation to find pressure and using the formula for integration of 1/x. They also consider the concept of differentiation and definite integration. The conversation concludes with a mention of finding the pressure at a specific height and the solution of the problem being continued at a later time.

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Raghav Gupta

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How you can write n as?

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goonking

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n? number of moles?Raghav Gupta said:How you can write n as?

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Raghav Gupta

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Yesgoonking said:n? number of moles?

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Raghav Gupta

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n = mass/molecular mass

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Raghav Gupta

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The approach is right. What do you get if you integrate?goonking said:## Homework Statement

View attachment 83508

## Homework Equations

PV=nRT

## The Attempt at a Solution

not sure if this is the right approach

View attachment 83509

plugging into -ρg gives us -PMg/RT = dP/dy

now we have to integrate both sides to find P?

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goonking

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well, i have no idea how to integrate that! :(Raghav Gupta said:The approach is right. What do you get if you integrate?

i only know how to integrate stuff like x

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Raghav Gupta

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No, problem.goonking said:well, i have no idea how to integrate that! :(

i only know how to integrate stuff like x^{2}+ 3

Do you know differentiation?

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goonking

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no, I'm suppose to take that next year :(Raghav Gupta said:No, problem.

Do you know differentiation?

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Raghav Gupta

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Oh, okaygoonking said:no, I'm suppose to take that next year :(

Then for the moment remember

$$ \int dx/x = lnx + C $$

Now use this in your problem.

And tell what you are getting.

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goonking

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i'm staring at this and still have no idea what to do, ok, so I know i can take the constants out and put it behind the integralRaghav Gupta said:Oh, okay

Then for the moment remember

$$ \int dx/x = lnx + C $$

Now use this in your problem.

And tell what you are getting.

(-Mg/RT) ∫ P = dP/dy

correct? all the constants are out except P

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Raghav Gupta

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$$ \frac{-Mg}{RT}\int_0^{8812} dy = \int_{10^5}^P \frac{dp}{P} $$

At ground height is zero and pressure 10

At height 8812 m we have to find pressure. The limits are taken accordingly.

Now I guess you know how to solve further ?

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goonking

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the left side of the equation should equal -1.095, correct?Raghav Gupta said:

$$ \frac{-Mg}{RT}\int_0^{8812} dy = \int_{10^5}^P \frac{dp}{P} $$

At ground height is zero and pressure 10^{5}pascals.

At height 8812 m we have to find pressure. The limits are taken accordingly.

Now I guess you know how to solve further ?

- #14

goonking

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I must go to school now, i will finish this problem next time.goonking said:the left side of the equation should equal -1.095, correct?

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Raghav Gupta

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Yes, and what right side evaluates to?goonking said:the left side of the equation should equal -1.095, correct?

The ideal gas law is a mathematical equation that describes the relationship between the pressure, volume, temperature, and number of moles of an ideal gas. In terms of density, the ideal gas law can be written as **ρ = (MP)/(RT)**, where ρ is the density, M is the molar mass, P is the pressure, R is the gas constant, and T is the temperature.

In the ideal gas law, density is directly proportional to pressure and molar mass, and inversely proportional to temperature and volume. This means that as pressure or molar mass increases, density increases, while as temperature or volume increases, density decreases.

The ideal gas law is important because it allows us to predict the behavior of gases under different conditions. In terms of density, it helps us understand how the density of a gas changes with variations in pressure, volume, temperature, and number of moles.

The ideal gas law is used in many practical applications, such as in the design of gas storage tanks, internal combustion engines, and refrigeration systems. It is also used in industries like chemical engineering, where the behavior of gases is critical in the production of various products.

The ideal gas law assumes that gases behave ideally, which means they have no intermolecular forces and occupy no volume. In reality, gases do have intermolecular forces and occupy some volume, especially at high pressures and low temperatures. Therefore, the ideal gas law is not accurate for all gases under all conditions, and other equations, such as the van der Waals equation, must be used to account for these deviations from ideality.

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