Thermodynamics atmospheric pressure Question

[benjibutton]

Homework Statement

A liter of air, initially at room temperature and atmospheric pressure, is heated at constant pressure until it doubles in volume. Calculate the increase in its entropy during this process.

so T_i= 300K, Volume which is 2V_i=V_f; Pressure is constant

Homework Equations

ΔS @constant pressure is = ∫C_p/T dT (where T_i -> T_f

The Attempt at a Solution

so since I don't know what T is but I know how the volume changes is there a way I can relate it? is it just PV=nkT so use V as an analog to gauge the proportional change in T? so would that give me something like Cp*Ln[2]?

 

[Simon Bridge]

But you do know what T is - it's the temperature. You mean you are not told what the final temperature is?
Then - yes - you would need to draw information in from another source - the equation you are using for entropy change is for an ideal gas, so it seems reasonable to use the ideal gas law to relate temperature and volume. Unless you have some notes about how air behaves?

http://www.thebigger.com/chemistry/...anges-of-an-ideal-gas-in-different-processes/

 

[benjibutton]

Yeah, sorry; the final T is unknown. So do I write it as Ln[T_f/T_i]? where T_f= 2PV_i/Nk ? how do I solve for N?

 

[benjibutton]

OK, I think I figured it out. I got it to be T_f=600K so it becomes 5/2*R*Ln(600K/300K)

 

[rude man]

OK, I think I figured it out. I got it to be T_f=600K so it becomes 5/2*R*Ln(600K/300K)

What happened to n? Your answer is for 1 mole, not 1 liter.

 

[benjibutton]

wouldn't n be the same for both sides of the ideal gas law, which would lead them to cancel out?

 

[benjibutton]

I also never had n, so I assumed you hold is constant, which would still give me the same result, since all other variables are held constant. Unless there's something else I'm missing.

 

[Simon Bridge]

At constant pressure $$\frac{PV_f=nRT_f}{PV_i=nRT_i}\implies \frac{V_f}{V_i} = \frac{T_f}{T_i}$$ ... it's a law that got named after someone.
It means you don't actually have to calculate the temperatures here if you know the volumes.

Of course you can also look up the molar density of air "at room temperature and atmospheric pressure".

 

[rude man]

wouldn't n be the same for both sides of the ideal gas law, which would lead them to cancel out?

Does it make sense to you that the change in entropy of 1 mole of gas is the same as the change in 1 liter? Your formula gave the change for 1 mole. So figure out how many moles in 1 liter of air at STP.

Your formula for ΔS = C_p ln(T2/T1) is correct. But C_p = nc_p and you only have c_p. BTW air is essentially a diatomic gas for which c_p ~ (7/2)R.