During an experiment, an ideal gas is found to obey an additional law pV^{2} = 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

PV^{2} = c

V^{2} 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 T_{2} and T_{1} , we get

⇒ \[\int\limits_{V_1}^{2V} = {-}\int\limits_{T_1}^{T_2}\]

⇒ ln( 2V) - ln(V) = ln (T_{1}) - ln (T_{2})

⇒ `ln ((2V)/V)` = `ln ((T_1)/(T_2))`

⇒ `T_2 = T_1/2`