Question

Initial temperature of an ideal gas is 75 °C. At what temperature, the sample of neon gas would be heated to double its pressure, if the initial volume of gas is reduced by 15%?
(a) 319 °C
(b) 592 °C
(c) 128 °C
(d) 60 °C

Answer 1

Hint: In order to solve the question we need to understand what an ideal gas is. Ideal gas is a theoretical or hypothetical gas with no intermolecular or interparticle force of attraction thus the internal potential energy of the gas is zero and has only the kinetic energy. In case of ideal gas the size of particles is very small and elastic collision occurs between the particles.

Complete answer:
The Ideal gas Equation is given as-
$PV = nRT - - - - (1)$
Where,
$P$ is Pressure of the gas;
$V$is the Volume of gas;
$T$is the temperature of the gas;
$n$ is number of moles of gas;
$R$is gas constant whose value is 8.314J/moleK
This Equation is derived from-
Boyle’s Law- Volume and Pressure of the gas are inversely related to one another at constant temperature and number of moles.
$V\alpha \dfrac{1}{P}$
We can write in terms of initial and final conditions as-
${P_1}{V_1} = {P_2}{V_2} - - - - (2)$
Avogadro’s Law- Volume and number of moles of gas are directly proportional to each other at a constant temperature and pressure.
$V\alpha n$
We can write in terms of initial and final conditions as-
$\dfrac{{{V_1}}}{{{n_1}}} = \dfrac{{{V_2}}}{{{n_2}}} - - - - (3)$
Charles’s law- Volume and temperature of the gas are directly proportional to each other at constant pressure and number of moles.
$V\alpha T$
We can write in terms of initial and final conditions as-
$\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}} - - - - (4)$
On rearranging Equation (1) we have,
$\dfrac{{PV}}{{nRT}} = Z = 1 - - - - (5)$ [Value in case of ideal gas]
Where $Z$ is called the Compressibility factor whose value is 1 in case of ideal gas and deviates largely from 1 in case of real gas.
The equation can also be written as-
$\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}} - - - - (6)$
Given in the question we have,
Initial temperature$({T_1}) = 75^\circ C = 75 + 273 = 348K$
Pressure is doubled so ${P_2} = 2{P_1}$
If initial volume is suppose ${V_1}$ so the final volume would be
${V_2} = {V_1} - \dfrac{{15}}{{100}}{V_1} = {V_1} - 0.15{V_1} = 0.85{V_1}$
We have to calculate the final temperature. Using Equation (6),
$\dfrac{{{P_1}{V_1}}}{{348}} = \dfrac{{2{P_1} \times 0.85{V_1}}}{{{T_2}}}$
On solving this Equation we get, ${T_2} = 591.6K = 591.6 - 273 = 318.6\alpha 319^\circ C$

So option (a) is correct.

Note:
In the real life conditions the Ideal gas condition is shown at only High temperature and pressure conditions. We have real gases mostly in actual situations with force of attraction between the interparticles. In the case of real gas the ideal gas equation is modified to give the ‘Van der waals gas Equation’.