Question

During an experiment, an ideal gas is found to obey an additional law pV² = constant. The gas is initially at a temperature T and volume V. Find the temperature when it expands to a volume 2V.

Use R = 8.3 J K^-1 mol^-1

Answer 1

Applying equation of state of an ideal gas, we get
PV = nRT
⇒ P = \[\frac{nRT}{V}  . . . 1 \]
Taking differentials, we get
⇒ PdV + VdP = nRdT  . . . 2
Applying the additional law, we get
PV² = c
V² dP + 2VPdV = 0
⇒ VdP + 2PdV = 0  . . . 3
Subtracting eq. (3) from eq. (2) , we get
 PdV = -nRdT
⇒ dV = \[\ {-}\frac{nR}{P}dT \]
Now ,
⇒ dV = \[\ {-}\frac{V}{T}dT \]  [From  eq. (1}]
⇒ \[\frac{dV}{V} = {-}\frac{dT}{T} \] 
Integrating between T₂ and T₁ , we get
⇒ \[\int\limits_{V_1}^{2V} = {-}\int\limits_{T_1}^{T_2}\]
⇒  ln( 2V) - ln(V) = ln (T₁) - ln (T₂) 
⇒ `ln ((2V)/V)` = `ln ((T_1)/(T_2))`
⇒ `T_2 = T_1/2`