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Question: One day, when the temperature and pressure were \[300{\text{K}}\] and \[760{\text{mm}}\] , a mass of...

One day, when the temperature and pressure were 300K300{\text{K}} and 760mm760{\text{mm}} , a mass of gas had a volume of 1200mL1200{\text{mL}} . On the next day, the volume had changed to 1218mL1218{\text{mL}} while the pressure was the same. What was the temperature on the next day?
A.546K546{\text{K}}
B.304.5K304.5{\text{K}}
C.31.5K31.5{\text{K}}
D.300K300{\text{K}}

Explanation

Solution

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 V1V2=T1T2\dfrac{{{{\text{V}}_1}}}{{{{\text{V}}_2}}} = \dfrac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} .
Formula Used: Ideal gas equation is: PV=nRT{\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: moles=mass(gram)molar mass(g/mol){\text{moles}} = \dfrac{{{\text{mass}}\left( {{\text{gram}}} \right)}}{{{\text{molar mass}}\left( {{\text{g}}/{\text{mol}}} \right)}}, universal gas constant is given as: R=8.31joulemoleK=2CalmoleK=0.082LatmmoleK{\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 P1P2=V2V1\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 V1V2=T1T2\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 P1P2=T1T2\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, V1=1200mL{{\text{V}}_1} = 1200{\text{mL}} , V2=1218mL{{\text{V}}_2} = 1218{\text{mL}} and T1=300K{{\text{T}}_1} = 300{\text{K}} .
In order to find T2{{\text{T}}_2} , we can use V1V2=T1T2\dfrac{{{{\text{V}}_1}}}{{{{\text{V}}_2}}} = \dfrac{{{{\text{T}}_1}}}{{{{\text{T}}_2}}} .
Therefore, 12001218=300T2\dfrac{{1200}}{{1218}} = \dfrac{{300}}{{{{\text{T}}_2}}} and on calculating:
T2=304.5K{{\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.