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Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: cofi on April 19, 2006, 07:18:25 AM

Title: work done by the gas?
Post by: cofi on April 19, 2006, 07:18:25 AM
1. A sample of carbon dioxide of mass 2,45g at 27°C is allowed to expand reversibly and adiabatically from 500 cm3 to 3.00 dm3. What is the work done by the gas?

2. A sample of 4.0 mol O2 is originally confined in 20 dm3 at 270 K and then undergoes adiabatic expansion against a constant pressure of 600 Torr until the volume has increased by a factor of 3.0. Calculate q, w, deltaT, deltaU and deltaH. (The final pressure of the gas is not necessarily 600 Torr.)


I've been trying to solve these problems for about 2 hours... I just can't find the right formula to start with...
Title: Re: work done by the gas?
Post by: Hunt on April 19, 2006, 03:55:30 PM
In this case dU = -dW

Remember that for an adiabatic process,? T1V1(gamma - 1) = T2V2(gamma - 1)

You can determine the final temp from here ( I think gamma for CO2 is given in your thermodynamics book ), and then find dU.
Title: Re: work done by the gas?
Post by: plu on April 27, 2006, 09:02:29 AM
Quote from: Vant_Hoff on April 19, 2006, 03:55:30 PM
In this case dU = -dW

I believe this should be dU = dW.  Then, a decrease in temperature would result in a negative sign for work, indicating work being done by the system through expansion.
Title: Re: work done by the gas?
Post by: Hunt on April 27, 2006, 06:00:50 PM
dU = -dW is correct taking into consideration the work done by the gas on the surr.

dWint = - dWext

dWext = - PdV , dWint = PdV

( this can be proven using vectors )

dU = -dWint + dq ---> dU = - dWint = - PdV

dU = dWext +dq ---> dU = dWext = - PdV

The expression of dU depends on how u define the work ... but it's the same value.
Title: Re: work done by the gas?
Post by: Donaldson Tan on April 27, 2006, 07:05:13 PM
cofi: what data are you provided?