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

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: ${\text{1 atm = 101325 pascals}}$