Question

The temperature of a gas is $T$ K. What would be the temperature \[6\] volume and pressure, both will reduce to half of the initial values?
A.\[T/2.\]
B.\[T/4.\]
C.\[T/3.\]
D.\[T/8.\]

Answer 1

Hint: We know that the given solution will be solved by using the ideal equation (An ideal equation is the relation between pressure, Temperature, amount of substance and volume). The ideal equation is a combination of Boyle’s law, Charles’s law and Avogadro’s law.

Complete answer:
As we know, the Gaseous state is a less ordered state where the molecules are separated from each other by large distances and have weak intermolecular forces. The gases have all the three types of motion–translator, rotatory and vibratory and possess high kinetic energy. An ideal gas is a theoretical gas composed of many randomly moving point particles that are not subject to interparticle interactions. The ideal gas concept is useful because it obeys the ideal gas law.
The four important laws we need to mesmerize are; Boyle’s law: It states that at constant temperature and amount of gas pressure is inversely proportional to the volume of gas. Charles’s law: It states that at constant pressure an amount of gas volume is directly proportional to the temperature of gas. Avogadro’s law: It states that at constant pressure and temperature the amount of gas is directly proportional to volume of gas. Gay Lussac’s law: It states that at constant volume of gas the pressure of gas is directly proportional to its temperature. From the gas equation, various other relationships can be obtained like Boyle's law, Charles’s law and Avogadro's law.
Here we have, \[{{P}_{1}}=P,~~{{T}_{1}}=T,~~{{V}_{1}}=V\]
These can be re-written as \[{{P}_{2}}=\dfrac{P}{2},~~{{V}_{2}}=\dfrac{V}{2},~~{{T}_{2}}=?\]
By using the ideal gas equation we get; $\dfrac{{{P}_{1}}{{V}_{1}}}{{{T}_{1}}}=\dfrac{{{P}_{2}}{{V}_{2}}}{{{T}_{2}}}$
For R.H.S we get; $\dfrac{P\times V}{T}=\dfrac{P\times V}{2\times 2\times {{T}_{2}}}$
On further solving we get; ${{T}_{2}}=\dfrac{T}{4}.$
Therefore, the correct answer is option B.

Note:
Remember that the temperature obtained from the gas equation is converted from the standard unit of Kelvin to Celsius. Also, the ideal gas equation used, usually predicts the approximate behaviour of the real gases. But, at low temperature and higher pressure, it deviates from the ideal gas behaviour.