Question

The intercept on y-axis of the line drawn for log P (P in atm) and $ log \dfrac{1}{V}(V in litre) $ for 1 mole of an ideal gas at 27° C is equal to:
(A) log 2.463
(B) log 24.63
(C) log 22.4
(D) log 2.24

Answer 1

Hint: Plotting the Boyle’s equation in straight line equation form ( $ {\text{y = mx + c}} $ ) gives the intercept value for the y-axis. Since ideal gas obeys Boyle’s law and Charle’s law they do give a straight line with slope and intercept determined easily from the respective equation.

Complete step by step solution:
According to Boyle’s law, the pressure (P) of a given mass of an ideal gas is inversely proportional to its volume (V) at constant temperature. This law can be mathematically expressed as follows:
$ {{P\alpha V}} \Rightarrow {\text{PV = K}} $
$ {\text{logP = log(}}\dfrac{{\text{1}}}{{\text{V}}}{\text{) + logK}} $
Comparing this to the straight-line equation, $ {\text{y = mx + c}} $ , where m is the slope and c is the y- intercept.
In Boyle’s law, when log P and $ log \dfrac{{\text{1}}}{{\text{V}}} $ is plotted, a straight line with y- intercept equal to K is obtained.
For an ideal gas, $ {\text{PV = nRT}} $ , it is given that 1 mole of ideal gas at a temperature of 27° C equal to 300 K.
Where R is the gas constant whose value is 0.0821 L atm mol-1 K-1 and T is the temperature expressed in kelvin (K).
We get, $ {\text{PV = RT}} $ . Comparing this with Boyle’s law, we get, $ {\text{K = RT}} $
 $
   \Rightarrow {\text{K = 0}}{\text{.0821 x 300 }} \\
   \Rightarrow {\text{K = 24}}{\text{.63}} \\
 $
Thus, y-intercept of Boyle’s law is log K= log 24.63
So, the correct option is (B).

Additional Information:
Charles’s law states that volume of an ideal gas at constant pressure is proportional to its absolute temperature.
 $ {{V\propto T}} $

Note:
According to the ideal gas equation, the standard temperature and pressure should be 273 K and 1 atm respectively.
In case of temperature given in Celsius it can be converted to kelvin by following expression: $ {{1^\circ C = 0 + 273 K}}{\text{.}} $
In case of pressure given in pascals it can be converted to atm by following expression: $ {\text{1 atm = 101325 pascals}} $