How Does Air Temperature and Pressure Affect Density Calculations?

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The discussion focuses on calculating the temperature of dry air at an airport based on a measured station pressure of 1014 mb and a known density of 1.3 kg/m^3. The Ideal Gas Law is applied, with the formula P = ρRT, where R is the gas constant for dry air. After determining the initial temperature, the conversation shifts to recalculating the air density when the surface pressure decreases to 1010 mb while keeping the temperature constant. Participants emphasize the importance of unit consistency in these calculations. The thread highlights the application of thermodynamic principles to understand how air temperature and pressure affect density.
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A weather station at the airport measured a station pressure of 1014 mb. The density of dry air is 1.3 kg/m^3. The gas constant for dry air, R, is 287 J/kg-K. Calculate the temperature of the dry air at the airport.

1 mb = 100 J/m ^3

AND

The surface pressure at the airport then decreased to 1010 mb but the air temperature reamined the same (as the answer in #1). Calculate new density of air.*I'm really confused and I would greatly appreciate it if someone can walk me through this problemThis is what I have so far, I know you have to use the Ideal Gas law
P= p R T

1014 mb = 1.3 kg/m^3 x 287 j/kg-k x T
 
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Hey, that looks really good so far! Surely you know enough algebra to take it from there. Just as a comment though, make sure you watch your units!
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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