Question

One day, when the temperature and pressure were \[300{\text{K}}\] and \[760{\text{mm}}\] , a mass of gas had a volume of \[1200{\text{mL}}\] . On the next day, the volume had changed to \[1218{\text{mL}}\] while the pressure was the same. What was the temperature on the next day?
A.\[546{\text{K}}\]
B.\[304.5{\text{K}}\]
C.\[31.5{\text{K}}\]
D.\[300{\text{K}}\]

Answer 1

Hint: The process in which pressure remains the same for a given amount of gas is known as an isobaric process. For isobaric process, Volume and temperature are related as \[\dfrac{{{{\text{V}}_1}}}{{{{\text{V}}_2}}} = \dfrac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}}\] .
Formula Used: Ideal gas equation is: \[{\text{PV}} = {\text{nRT}}\], where P is pressure of the gas, V is the volume of gas, n is number of moles of the gas, R is universal gas constant and T is the temperature. Number of moles can be given as: \[{\text{moles}} = \dfrac{{{\text{mass}}\left( {{\text{gram}}} \right)}}{{{\text{molar mass}}\left( {{\text{g}}/{\text{mol}}} \right)}}\], universal gas constant is given as: \[{\text{R}} = 8.31\dfrac{{{\text{joule}}}}{{{\text{mole}} - {\text{K}}}} = 2\dfrac{{{\text{Cal}}}}{{{\text{mole}} - {\text{K}}}} = 0.082\dfrac{{{\text{L}} - {\text{atm}}}}{{{\text{mole}} - {\text{K}}}}\].

Complete step by step answer:
If temperature remains the same for a given amount of gas then according to the ideal gas equation, pressure of that gas will be inversely proportional to its volume. For this relation of pressure and volume can be given by \[\dfrac{{{{\text{P}}_1}}}{{{{\text{P}}_2}}} = \dfrac{{{{\text{V}}_2}}}{{{{\text{V}}_1}}}\] .
Similarly, if pressure remain same for a given amount of gas then according to ideal gas equation, volume and temperature of that gas can be related as \[\dfrac{{{{\text{V}}_1}}}{{{{\text{V}}_2}}} = \dfrac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}}\] .
And if volume remain same for a given amount of gas, then its pressure and temperature can be related as \[\dfrac{{{{\text{P}}_1}}}{{{{\text{P}}_2}}} = \dfrac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}}\] .
According to given data in question;
Pressure remain same for given amount of gas, \[{{\text{V}}_1} = 1200{\text{mL}}\] , \[{{\text{V}}_2} = 1218{\text{mL}}\] and \[{{\text{T}}_1} = 300{\text{K}}\] .
In order to find \[{{\text{T}}_2}\] , we can use \[\dfrac{{{{\text{V}}_1}}}{{{{\text{V}}_2}}} = \dfrac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}}\] .
Therefore, \[\dfrac{{1200}}{{1218}} = \dfrac{{300}}{{{{\text{T}}_2}}}\] and on calculating:
\[{{\text{T}}_2} = 304.5{\text{K}}\] .

Thus, the correct option is B.

Note:
 A gas which follows all gas laws and gas equations at every possible temperature and pressure is known as ideal or perfect gas. Potential energy of an ideal gas is taken to be zero. Its internal energy is directly proportional to absolute temperature. All real gas behaves as ideal gas at high temperature and low pressure. Internal energy of real gas depends upon temperature, pressure and volume.