Question

The critical pressure $ {{P}_{C}} $ and critical temperature $ {{T}_{C}} $ for a gas obeying Van Der Waals equation are $ 80 $ atm and $ 87{}^\circ C. $ Molar mass of the gas is $ 130g/mole. $ The compressibility factor for the above gas will be smaller than unity under the following conditions:
(A) $ 1 $ atm and $ 800{}^\circ C $
(B) $ 1 $ atm and $ 1200{}^\circ C $
(C) $ 1 $ atm and $ 1000{}^\circ C $
(D) $ 1 $ atm and $ 1100{}^\circ C $

Answer 1

Hint :We know that the compressibility factor $ \left( Z \right) $ is a correction factor which describes the deviation of a real gas from its behaviour of an ideal gas. It is simply the ratio of molar volume of the gas to the molar volume of an ideal gas subjected to the same identical temperature and pressure.

Complete Step By Step Answer:
As we know that the critical temperature is defined as the temperature of a substance in its critical state beyond which it cannot be liquefied. Critical pressure of a fluid is defined as the vapour pressure of the fluid at the critical temperature. Critical volume is the volume occupied by the fluid or substance in its critical state, critical pressure and temperature. The gases can be liquefied by compressing the gas at a suitable temperature. The liquefaction of gases becomes difficult as the temperature of the gases increases. The increase in temperature increases the kinetic energy of gas particles. The critical temperature of a gas is a temperature at and above which the vapour of the substance cannot be liquefied irrespective of the pressure applied on the substance. The pressure which is needed to liquefy the gas at critical temperature is known as critical pressure and volume as critical volume. Here we have $ {{T}_{C}}=87{}^\circ C=360K $ and $ {{P}_{C}}=80 $ atm.
Thus, $ Z=\dfrac{PV}{nRT}=1 $ when the repulsive forces and attractive forces between gaseous balanced out each other then we say that compressibility factor is unity;
At $ {{T}_{B}}\Rightarrow z=1 $ and at $ T<{{T}_{B}}\Rightarrow zThus, the Boyle temperature is given out by $ {{T}_{B}}=\dfrac{27}{8}\times {{T}_{C}} $
On further substitution we get; $ {{T}_{B}}=\dfrac{27}{8}\times 360=1215K~~~~or~~~~~~~942{}^\circ C $
 $ {{T}_{B}}=942{}^\circ C\approx 800{}^\circ C $
Therefore, the correct answer is option A.

Note :
Remember that in the ideal gas equation there is no force of attraction between the gas molecules. But this is not observed in real life. Hence, the correction factor is introduced in the van der Waal equation. Using the ideal gas equation we will get the value of pressure. Equating the equation we will find the value of one of the constants.