- #1

speny83

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## Homework Statement

A sample of carbon dioxide, mass 2.45g, is allowed to expand

__reversible & adiabaticly__from 500cm

^{3}to 3.00 dm

^{3}

What is the work done by the gas

## Homework Equations

dw=-p

_{ex}dV

du=dq+dw

du=nC

_{v,m}dt

## The Attempt at a Solution

At first i just wanted to say that because dw=-p

_{ex}dV and p

_{ex}=p=[itex]\frac{nrt}{v}[/itex] so w=-nrt∫dV/V....w=-nrt*ln(v

_{f}/v

_{i})

however this doesent give me the correct sollution, but i dont understand why i cant say that.

my second thought was to say that because du=dq+dw and dq=0 for adiabatics that it would make du=dw and with du=nC

_{v,m}dt so w=nC

_{v,m}ΔT i could go about it this way, but i need to find the final temp, as an aside: is the reason i cannot use charles law here because work is being done so p is not constant?

ok so i need final T wich i can get by T

_{f}=([itex]\frac{Vf}{Vi}[/itex])

^{1/c}

my notes say that c=Cvm/R and, this is not part of the notes, Cvm for CO2 (because its linear) is just 5R/2 right?

however the solution guide goes into all this c=[itex]\frac{Cvm}{R}[/itex]=[itex]\frac{Cpm-R}{R}[/itex]=3.463

but they dont explain any numbers...if you try to reverse calc it (3.463*8.3145)+8.3145 you get 37 somethig which doesent coordinate to anything either....also wouldnt Cp,m just be 7R/2?

now im so confused i dont know what to do

I moved on and worked some other problems (involving CO2) where i had to find [itex]\gamma[/itex] being Cpm/Cvm doing this using Cpm=7R/2 and Cvm=5R/2 worked just fine so I am inclined to believe that the above just doesent make a lick of sense

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