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
Solution
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`
