Question

The volume of a gas at ${20^0}C$ is $200ml$. If the temperature is reduced to $ - {20^\circ}C$ at constant pressure, its volume will be
(A) $172.6ml$
(B) $17.26ml$
(C) $192.7ml$
(D) $19.27ml$

Answer 1

Hint: First apply the ideal gas law, and then use the condition of constant pressure in that equation. After that put all the values provided in the question in that equation and get the required answer. At constant pressure, we know that volume divided by the temperature of the gas is also constant.

Formula used:
$PV = nRT$
 $\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$


Complete answer:
According to the ideal gas law, the product of the pressure and volume of one gram of an ideal gas is equal to the product of the gas's absolute temperature and the universal gas constant.
By Ideal Gas Law, we know;
$PV = nRT$
Where P is pressure
V is volume
R is the universal gas constant
T is temperature
At constant pressure,
$\dfrac{{nR}}{P} = \dfrac{V}{T} = constant$
Therefore, $\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$ (equation 1)
From the question, we know;
${V_1} = 200ml$
${V_2} = ?$
${T_1} = {20^0}C = 293K$
${T_2} = - {20^0}C = 253K$
Substituting all the values in equation 1, we get;
$\dfrac{{200ml}}{{293K}} = \dfrac{{{V_2}}}{{253K}}$
By solving, we get;
${V_2} = 172.6ml$ (which is the final volume)
The correct option is (A).



Note: This question can be solved directly using Charle’s law which states that When the pressure on a dry gas sample is kept constant, the Kelvin temperature and volume are in direct proportion, i.e., $\dfrac{{{V_2}}}{{{V_1}}} = \dfrac{{{T_2}}}{{{T_1}}}$. Make sure to convert the temperature unit into kelvin, otherwise, it can cause an error in the solution.