
When two moles of an ideal gas \[\left( {{C_{p,m}} = \dfrac{5}{2}R} \right)\] is heated from $300K$ to $600K$ at constant pressure, the change in entropy of the gas ($\Delta S$ ) is:
A. $\dfrac{3}{2}R\ln 2$
B. $3R\ln 2$
C. $5R\ln 2$
D. $\dfrac{5}{2}R\ln 2$
Answer
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Hint: The second law of thermodynamics states that entropy in an isolated system which is the combination of a subsystem under study and its surroundings, increases during all spontaneous chemical and physical processes. The Clausius equation of $\dfrac{{d{q_{rev}}}}{T} = \Delta S$ introduces the measurement of entropy change, $\Delta S$ .
Complete step by step answer:
Entropy change describes the direction and quantifies the magnitude of simple changes such as heat transfer between systems – always from hotter to cooler spontaneously. The thermodynamic entropy therefore has the dimension of energy divided by temperature, and the unit joule per Kelvin ($J/K$ ) in the International System of Units (SI).
The quantity of heat required to raise the temperature of one mole of gas through \[1K\] (or\[1^\circ C\] ) when pressure is kept constant is called molar specific heat at constant pressure. It is denoted by\[{C_{p,m}}\] . Its S.I. unit is $Jmo{l^{ - 1}}{K^{ - 1}}$.
The relation between the change in entropy of a system and the molar specific heat is given as:
$\Delta S = n{C_{p,m}} \times \ln \left( {\dfrac{{{T_2}}}{{{T_1}}}} \right)$
Where, $\Delta S = $ change in entropy
$n = $ number of moles = 2
${T_1} = 300K$
${T_2} = 600K$
\[{C_{p,m}} = \dfrac{5}{2}R = \]molar specific heat at constant pressure
Substituting these values in the above equation, we have:
$\Delta S = 2 \times \dfrac{5}{2}R \times \ln \left( {\dfrac{{600}}{{300}}} \right) = 5R\ln 2$
Thus, the correct option is C. $5R\ln 2$ .
Note:
Thermodynamic entropy is an extensive property, meaning that it scales with the size or extent of a system. In many processes it is useful to specify the entropy as an intensive property independent of the size, as a specific entropy characteristic of the type of system studied. Specific entropy may be expressed relative to a unit of mass, typically the kilogram ($Jk{g^{ - 1}}{K^{ - 1}}$ ). Alternatively, in chemistry, it is also referred to one mole of substance, in which case it is called the molar entropy with a unit of $Jmo{l^{ - 1}}{K^{ - 1}}$.
Complete step by step answer:
Entropy change describes the direction and quantifies the magnitude of simple changes such as heat transfer between systems – always from hotter to cooler spontaneously. The thermodynamic entropy therefore has the dimension of energy divided by temperature, and the unit joule per Kelvin ($J/K$ ) in the International System of Units (SI).
The quantity of heat required to raise the temperature of one mole of gas through \[1K\] (or\[1^\circ C\] ) when pressure is kept constant is called molar specific heat at constant pressure. It is denoted by\[{C_{p,m}}\] . Its S.I. unit is $Jmo{l^{ - 1}}{K^{ - 1}}$.
The relation between the change in entropy of a system and the molar specific heat is given as:
$\Delta S = n{C_{p,m}} \times \ln \left( {\dfrac{{{T_2}}}{{{T_1}}}} \right)$
Where, $\Delta S = $ change in entropy
$n = $ number of moles = 2
${T_1} = 300K$
${T_2} = 600K$
\[{C_{p,m}} = \dfrac{5}{2}R = \]molar specific heat at constant pressure
Substituting these values in the above equation, we have:
$\Delta S = 2 \times \dfrac{5}{2}R \times \ln \left( {\dfrac{{600}}{{300}}} \right) = 5R\ln 2$
Thus, the correct option is C. $5R\ln 2$ .
Note:
Thermodynamic entropy is an extensive property, meaning that it scales with the size or extent of a system. In many processes it is useful to specify the entropy as an intensive property independent of the size, as a specific entropy characteristic of the type of system studied. Specific entropy may be expressed relative to a unit of mass, typically the kilogram ($Jk{g^{ - 1}}{K^{ - 1}}$ ). Alternatively, in chemistry, it is also referred to one mole of substance, in which case it is called the molar entropy with a unit of $Jmo{l^{ - 1}}{K^{ - 1}}$.
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