Question

At NTP, the volume of a gas is found to be 273 ml. What will be the volume of this gas at 600 mm Hg and 273 \[^\circ C\] ?
A.391.8 ml
B.380 ml
C.644 ml
D.750 ml

Answer 1

Hint: NTP stands for Normal condition of Temperature and pressure. This is considered to the ideal constant value while understanding the nature of the external parameters in a particular experimental setting. The values of temperature and pressure at NTP are given to be 293 K and 1 atm respectively.This question can be solved using a combined gas equation.

Complete step by step answer:
Now, let us write down the data that has been given to us:
Let the initial and final recorded temperatures be \[{T_1}\] and \[{T_2}\] respectively. Similarly let the initial and final recorded pressures and volumes of the given gas be \[{P_1}\] , \[{P_2}\] and \[{V_1}\] , \[{V_2}\] respectively.
To tabulate this data:
 \[{T_1}\] = 293 K
 \[{T_2} = {273^\circ }C\] \[ = \left( {273 + 273} \right)K = 546{\text{ }}K\]
\[{P_1}\] = 1 atm = 760 mm of Hg
\[{P_2}\] = 600 mm of Hg
 \[{V_1}\] = 273 ml
 \[{V_2}\] = ‘x’ ml = ?
To solve this question, we are going to use a formula named the combined gas equation. This equation helps in establishing a relationship between the values of temperature, pressure and volume of a given gas at two different readings. The mathematical representation of this equation can be given by:
 \[\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}\]
Rearranging the above equation:
 \[{V_2} = \dfrac{{{P_1}{V_1}}}{{{T_1}}}.\dfrac{{{T_2}}}{{{P_2}}}\]
Substituting the values form the given data in the above equation:
x = \[\dfrac{{760x273}}{{293}}.\dfrac{{546}}{{600}}\]
x = 644 ml
hence, the volume of the gas has been found to be 644 ml.

Hence, Option C is the correct option.

Note:

The combined gas equation is obtained from three different equations. It utilises the relations established in Boyle’s Law, Charles’ Law and Gay-Lussac law. Since it combines all these three relations, it is known as the Combined Gas equation.