9) In Example 5 on page 776, why should we have expected the coefficient of dT t
ID: 2261693 • Letter: 9
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9) In Example 5 on page 776, why should we have expected the coefficient of dT to be negative? This question is about Calculus III 14.3 local linearity and the differential 776 Chapter Fourteen DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES The density (in g/cm3) of carbon dioxide gas CO2 depends upon its temperature T (in °C) and pressure P (in atmospheres). The ideal gas model for CO2 gives what is called the state equation: Example 5 0.5363P T273.15 Compute the differential dp. Explain the signs of the coefficients of dT and dP Solution The differential for = f(T, P) is 0.5363P 0.5363 dT + d.P The coefficient of dT is negative because increasing the temperature expands the gas (if the pressure is kept constant) and therefore decreases its density. The coefficient of dP is positive because in easing the pressure compresses the gas (if the temperature is kept constant) and therefore increases cr its density.Explanation / Answer
For ideal gas we have PV/T= constant ,{let k and k >0 as P,V,T are positive quantities)---------------(1)
(P is pressure, V is volume and T ( in Kelvin) is temperature.)
So when pressure is constant, from (1) we get V = p T ( p = k/P = constant )----------(2)
From (2) we can see that increase (or decrease) in T will lead to increase (or decrease) in V.
or positive change in T implies positive change in V
or equivalently
V2 - V1 = p ( T2 - T1) ---------------------------(Using 2)
If we take V2 = V1 + dt and T2 = T1 + dt
Then we get dV = p ( dt ) ------------------------------------(3)
Now density = mass /volume
So for constant mass m
increase (or decrease) in volume means mass/volume decreases (or increases)
Consequently density decreases (or increases)-------------------------(4)
So from (3) and (4) we conclude that change in temperature and change in density are opposite to each other.
Note that in the given equation of density if we substitute mass/volume in place of density, then we get the same equation as (1) above. The 273.15 term is because temperature here is in celsius instead of kelvin.
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