- #1

DottZakapa

- 239

- 17

A volume

DATA:

Following there is what I've solved so far:

P

T

V

P

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 :

T

The solution to this is :

W=∫ p⋅dV=2nRT

but I don't get how.

Could somebody help me with this?

Thanks

*V*contains_{A}*n*mole of a bi-atomic ideal gas, initially at temperature*T*. Burning an amount_{A}*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 (T^{2}/ V) constant. Assuming that no heat is wasted in the environment.DATA:

*n= 0.3 [mole]*;*R= 0.082 [litre*atm/(mole*K)]*;*V*_{A}=9 [litre] ; T_{A}=300 [K]; 1 [cal]=4.18[J]]Following there is what I've solved so far:

P

_{A}=(nRT_{A})/ V_{A}T

_{B}= 2T_{A}V

_{B}=(T_{B}^{2}⋅V_{A})/ T_{A}^{2}= 4V_{A}P

_{B}=(nR2T_{A})/ 4V_{A}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 :

T

^{2}/ V = T_{A}^{2}/ V_{A}The solution to this is :

W=∫ p⋅dV=2nRT

_{A}but I don't get how.

Could somebody help me with this?

Thanks

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