How Is Work Calculated in a Thermodynamic Process with Constant T2/V Ratio?

[DottZakapa]

A volume V_A contains n mole of a bi-atomic ideal gas, initially at temperature T_A. 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)] ;
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_B²⋅V_A)/ T_A²= 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 :

T² / V = T_A² / 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

 

[Chestermiller]

How are T, V, T_A, and V_A related?

In terms of T_A, V_A, and n, what is the initial pressure p_A?

What is the pressure when the volume is V?

Chet

 

[DottZakapa]

How are T, V, T_A, and V_A related?

In terms of T_A, V_A, and n, what is the initial pressure p_A?

What is the pressure when the volume is V?

Chet

I've edited the template

 

[Chestermiller]

Express p as a function of n, T_A, V_A and V. Then integrate.

Chet