Physical Chemistry Barometric Derivation

[jen333]

Physical Chemistry Barometric Derivation!

Homework Statement

Derive an expression for the barometric equation that takes the temperature dependency (T_o=T-az) into account. Where To=ground level temperature, T=temperature at an altitude, a=proportion constant, z=altitdue

Homework Equations

p=p_oe^-gMz/RT

The Attempt at a Solution

I know how to derive the barometric equation formula to get to ln(p/po) which i know has to be part of the final expression. however, I having trouble with the integral on the other side. should i integrate for dz or dT?

 

[Nino MMP]

And how to i account for the temperature dependency?p=poe-gMz/RTln(p/po) = -gM/RT * Integral(dz or dT)Since we are accounting for temperature, we need to integrate with respect to temperature. The equation for To=T-az implies that dT=dz/a. Therefore, we can substitute dT in the integral to get:ln(p/po) = -gM/RT * Integral(dT)= -gM/RT * Integral(dz/a)= -gM/aRT * Integral(dz)So the final expression for the barometric equation with temperature dependency is: p=poe-gMz/aRT

 

[Blueshift5]

I would suggest approaching this problem by first considering the ideal gas law, which relates pressure, volume, and temperature. We can use this law to derive the barometric equation, which describes how pressure changes with altitude in a column of air.

Starting with the ideal gas law, we can rearrange it to isolate pressure (p):

pV = nRT

p = nRT/V

Next, we can consider a column of air of height z, with a cross-sectional area of A. The volume of this column of air can be represented as V = Az, where A is the cross-sectional area and z is the height. We can also assume that the number of moles of air (n) remains constant.

Substituting these values into the ideal gas law, we get:

p = nRT/Az

Next, we can use the fact that density (ρ) is equal to mass (m) divided by volume (V). In this case, the mass of the column of air remains constant, so we can say:

ρ = m/V = constant

Therefore, we can rewrite the ideal gas law as:

p = ρRT/Az

Now, we need to account for the temperature dependency. We can do this by substituting T = To - az, where To is the ground level temperature and a is a proportionality constant, into the ideal gas law. This gives us:

p = ρR(To - az)/Az

Next, we can use the barometric equation to relate pressure at an altitude (p) to pressure at ground level (po):

p = poe^(-gz/RT)

Substituting this into our previous equation, we get:

poe^(-gz/RT) = ρR(To - az)/Az

Solving for p/po, we get:

p/po = ρR(To - az)/Az * e^(gz/RT)

Finally, taking the natural logarithm of both sides, we get the expression for the barometric equation that takes temperature dependency into account:

ln(p/po) = ln(ρR(To - az)/Az * e^(gz/RT))

Therefore, to answer your question about the integral, you would need to integrate for dz because it is the variable that is changing with altitude. I hope this helps in your derivation.