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Find Work thermodynamics

  • Thread starter DottZakapa
  • Start date
  • #1
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A volume VA contains n mole of a bi-atomic ideal gas, initially at temperature TA. Burning an amount M of methane, whose calorific power P (produced heat per unit mass while burning) is 13271 [kcal/kg], the temperature is slowly doubled, simultaneously expanding the volume in order to maintain the ratio (T2 / V) constant. Assuming that no heat is wasted in the environment.
DATA:
n= 0.3 [mole]; R= 0.082 [litre*atm/(mole*K)] ;
VA=9 [litre] ; TA=300 [K]; 1 [cal]=4.18[J]]

Following there is what i've solved so far:
PA=(nRTA)/ VA

TB= 2TA

VB=(TB2⋅VA)/ TA2= 4VA

PB=(nR2TA)/ 4VA

Considering that pressure and temperature aren't constant during the transformation, I'm not sure which value instead of pressure I have to insert in the integral in order to evaluate the work during the transformation.

W=∫ p⋅dV

I guess that I shall use this relation :

T2 / V = TA2 / VA

The solution to this is :

W=∫ p⋅dV=2nRTA

but I don't get how.
Could somebody help me with this?
Thanks
 
Last edited:

Answers and Replies

  • #2
20,245
4,265
How are T, V, TA, and VA related?

In terms of TA, VA, and n, what is the initial pressure pA?

What is the pressure when the volume is V?

Chet
 
  • #3
160
9
How are T, V, TA, and VA related?

In terms of TA, VA, and n, what is the initial pressure pA?

What is the pressure when the volume is V?

Chet
I've edited the template
 
  • #4
20,245
4,265
Express p as a function of n, TA, VA and V. Then integrate.

Chet
 

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