- #1
DottZakapa
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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
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: