Answer
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Hint: In order to calculate the entropy change of the given Argon gas, we must first have an idea about what an entropy is. Entropy is said to be the measure of disorderliness or randomness present in a molecule of a system.
Complete step by step answer:
Let us understand about entropy. Entropy is referred to as the measure of disorderliness or randomness present in a molecule of a system. Entropy is also said to be a thermodynamic function. It is also a state function that depends on the state of the system and not on the path it followed. Entropy is usually represented by the symbol S, but entropy in the standard state is represented by the symbol \[S^\circ \]. The SI unit of the entropy is found to be \[J{K^{ - 1}}mo{l^{ - 1}}\]. Entropy is said to be an extensive property.
- Let us move onto the problem given. Argon is one of the noble gases and it is a monatomic gas. We can find the entropy change by the following formula:
\[\Delta S = n{C_V}\ln \dfrac{{{T_2}}}{{{T_1}}} + nR\ln \dfrac{{{V_2}}}{{{V_1}}}\]……... (1)
where \[{C_V} = \dfrac{3}{2}R\]
$\rightarrow$ \[{T_1} = 25^\circ C = 298K\]
$\rightarrow$ \[{T_2} = 100^\circ C = 373K\]
$\rightarrow$ \[{V_1} = 0.50d{m^3}\]
$\rightarrow$ \[{V_2} = 100d{m^3}\]
n = 1
Therefore, we can write the equation (1) as
\[\Delta S = 1 \times \dfrac{3}{2}R\ln \dfrac{{373}}{{298}} + 1 \times R\ln \dfrac{{100}}{{0.50}}\]
$\rightarrow$ \[\Delta S = \dfrac{3}{2}R \times 0.224 + R \times 5.298\]
$\rightarrow$ \[\Delta S = 0.336R + 5.298R = 5.634R\]
$\rightarrow$ \[\Delta S = 5.634 \times 8.3J{K^{ - 1}}mo{l^{ - 1}}\]
$\rightarrow$ \[\Delta S = 46.84J{K^{ - 1}}mo{l^{ - 1}}\]
Therefore, the entropy change was found to be \[46.84J{K^{ - 1}}mo{l^{ - 1}}\].
Note: We have to remember that when there is greater disorder or randomness then the entropy will increase. Therefore, gases have greater randomness compared to liquid and solid hence the entropy is greater in gases.
The order of entropy is given as
Gas > liquid > Solid
Complete step by step answer:
Let us understand about entropy. Entropy is referred to as the measure of disorderliness or randomness present in a molecule of a system. Entropy is also said to be a thermodynamic function. It is also a state function that depends on the state of the system and not on the path it followed. Entropy is usually represented by the symbol S, but entropy in the standard state is represented by the symbol \[S^\circ \]. The SI unit of the entropy is found to be \[J{K^{ - 1}}mo{l^{ - 1}}\]. Entropy is said to be an extensive property.
- Let us move onto the problem given. Argon is one of the noble gases and it is a monatomic gas. We can find the entropy change by the following formula:
\[\Delta S = n{C_V}\ln \dfrac{{{T_2}}}{{{T_1}}} + nR\ln \dfrac{{{V_2}}}{{{V_1}}}\]……... (1)
where \[{C_V} = \dfrac{3}{2}R\]
$\rightarrow$ \[{T_1} = 25^\circ C = 298K\]
$\rightarrow$ \[{T_2} = 100^\circ C = 373K\]
$\rightarrow$ \[{V_1} = 0.50d{m^3}\]
$\rightarrow$ \[{V_2} = 100d{m^3}\]
n = 1
Therefore, we can write the equation (1) as
\[\Delta S = 1 \times \dfrac{3}{2}R\ln \dfrac{{373}}{{298}} + 1 \times R\ln \dfrac{{100}}{{0.50}}\]
$\rightarrow$ \[\Delta S = \dfrac{3}{2}R \times 0.224 + R \times 5.298\]
$\rightarrow$ \[\Delta S = 0.336R + 5.298R = 5.634R\]
$\rightarrow$ \[\Delta S = 5.634 \times 8.3J{K^{ - 1}}mo{l^{ - 1}}\]
$\rightarrow$ \[\Delta S = 46.84J{K^{ - 1}}mo{l^{ - 1}}\]
Therefore, the entropy change was found to be \[46.84J{K^{ - 1}}mo{l^{ - 1}}\].
Note: We have to remember that when there is greater disorder or randomness then the entropy will increase. Therefore, gases have greater randomness compared to liquid and solid hence the entropy is greater in gases.
The order of entropy is given as
Gas > liquid > Solid
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