Compressibility Factor Graph - Which gas attains a deeper minimum?
$begingroup$
As in the given graph, how do I predict which curve among that of hydrogen, methane, and carbon dioxide attains a deeper absolute minimum? (greater deviation from Z=1). In fact, how do I make this comparison for any gases for that matter, and on what factors does this depend?
Is there any temperature dependence of the above?
If nothing can be said in general, I'd at least like to know how to make the comparison for ammonia and methane, as that has previously been asked in a contest years ago.
I'm looking for more of qualitative comparison, but all sorts of answers are welcome.
physical-chemistry gas-laws
asked Jan 12 at 5:01
arya_starkarya_stark
1,415320
$endgroup$
add a comment |
$begingroup$
As in the given graph, how do I predict which curve among that of hydrogen, methane, and carbon dioxide attains a deeper absolute minimum? (greater deviation from Z=1). In fact, how do I make this comparison for any gases for that matter, and on what factors does this depend?
Is there any temperature dependence of the above?
If nothing can be said in general, I'd at least like to know how to make the comparison for ammonia and methane, as that has previously been asked in a contest years ago.
I'm looking for more of qualitative comparison, but all sorts of answers are welcome.
physical-chemistry gas-laws
asked Jan 12 at 5:01
arya_starkarya_stark
1,415320
$endgroup$
add a comment |
$begingroup$
As in the given graph, how do I predict which curve among that of hydrogen, methane, and carbon dioxide attains a deeper absolute minimum? (greater deviation from Z=1). In fact, how do I make this comparison for any gases for that matter, and on what factors does this depend?
Is there any temperature dependence of the above?
If nothing can be said in general, I'd at least like to know how to make the comparison for ammonia and methane, as that has previously been asked in a contest years ago.
I'm looking for more of qualitative comparison, but all sorts of answers are welcome.
physical-chemistry gas-laws
asked Jan 12 at 5:01
arya_starkarya_stark
1,415320
$endgroup$
As in the given graph, how do I predict which curve among that of hydrogen, methane, and carbon dioxide attains a deeper absolute minimum? (greater deviation from Z=1). In fact, how do I make this comparison for any gases for that matter, and on what factors does this depend?
Is there any temperature dependence of the above?
If nothing can be said in general, I'd at least like to know how to make the comparison for ammonia and methane, as that has previously been asked in a contest years ago.
I'm looking for more of qualitative comparison, but all sorts of answers are welcome.
physical-chemistry gas-laws
physical-chemistry gas-laws
asked Jan 12 at 5:01
arya_starkarya_stark
1,415320
asked Jan 12 at 5:01
arya_starkarya_stark
1,415320
asked Jan 12 at 5:01
arya_starkarya_stark
1,415320
asked Jan 12 at 5:01
arya_starkarya_stark
1,415320
asked Jan 12 at 5:01
arya_starkarya_stark
1,415320
1,415320
add a comment |
add a comment |
1 Answer
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There is a temperature dependence, for the same molecule the lower the temperature, the smaller is the minimum $Z$. You can see this if you plot $Z$ vs $P$ using the van der Waals equation. This makes sense because at lower temperature ( and hence lower energy) molecules are more able to associate with one another and so are less 'ideal', i.e. there is less energy with which to break up intermolecular interactions.
If you plot $Z$ vs $P$ using reduced pressure, temperature and volume, as $T_r=T/T_c,; P_r=P/P_c,;V_r=V/V_c$ where the subscript $c$ indicates values at the critical point then many gases fall onto the same plots at a given temperature. This is called the 'Law of corresponding states'. The constants $a$ and $b$ are eliminated from the final equation for $Z$, although it looks as thought they will be used, making the resulting equation 'universal'. Thus all gasses behave the same way when compared at corresponding conditions'
$Z$ vs $P_r$ at different reduced temperatures
For a van der Waals gas $(P+a/V_m^2)(V_m-b)=RT$ where $a$ and $b$ are obtained from experiment for each gas and $V_m$ is the molar volume and $T_C=8a/(27Rb), ; P_c=a/(27b^2) ,; V_{mc}=3b$ where $V_{mc}$ is critical molar volume.
answered Jan 12 at 10:02
porphyrinporphyrin
17.4k2954
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1 Answer
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1 Answer
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active
oldest
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active
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active
oldest
votes
$begingroup$
There is a temperature dependence, for the same molecule the lower the temperature, the smaller is the minimum $Z$. You can see this if you plot $Z$ vs $P$ using the van der Waals equation. This makes sense because at lower temperature ( and hence lower energy) molecules are more able to associate with one another and so are less 'ideal', i.e. there is less energy with which to break up intermolecular interactions.
If you plot $Z$ vs $P$ using reduced pressure, temperature and volume, as $T_r=T/T_c,; P_r=P/P_c,;V_r=V/V_c$ where the subscript $c$ indicates values at the critical point then many gases fall onto the same plots at a given temperature. This is called the 'Law of corresponding states'. The constants $a$ and $b$ are eliminated from the final equation for $Z$, although it looks as thought they will be used, making the resulting equation 'universal'. Thus all gasses behave the same way when compared at corresponding conditions'
$Z$ vs $P_r$ at different reduced temperatures
For a van der Waals gas $(P+a/V_m^2)(V_m-b)=RT$ where $a$ and $b$ are obtained from experiment for each gas and $V_m$ is the molar volume and $T_C=8a/(27Rb), ; P_c=a/(27b^2) ,; V_{mc}=3b$ where $V_{mc}$ is critical molar volume.
answered Jan 12 at 10:02
porphyrinporphyrin
17.4k2954
$endgroup$
add a comment |
$begingroup$
There is a temperature dependence, for the same molecule the lower the temperature, the smaller is the minimum $Z$. You can see this if you plot $Z$ vs $P$ using the van der Waals equation. This makes sense because at lower temperature ( and hence lower energy) molecules are more able to associate with one another and so are less 'ideal', i.e. there is less energy with which to break up intermolecular interactions.
If you plot $Z$ vs $P$ using reduced pressure, temperature and volume, as $T_r=T/T_c,; P_r=P/P_c,;V_r=V/V_c$ where the subscript $c$ indicates values at the critical point then many gases fall onto the same plots at a given temperature. This is called the 'Law of corresponding states'. The constants $a$ and $b$ are eliminated from the final equation for $Z$, although it looks as thought they will be used, making the resulting equation 'universal'. Thus all gasses behave the same way when compared at corresponding conditions'
$Z$ vs $P_r$ at different reduced temperatures
For a van der Waals gas $(P+a/V_m^2)(V_m-b)=RT$ where $a$ and $b$ are obtained from experiment for each gas and $V_m$ is the molar volume and $T_C=8a/(27Rb), ; P_c=a/(27b^2) ,; V_{mc}=3b$ where $V_{mc}$ is critical molar volume.
answered Jan 12 at 10:02
porphyrinporphyrin
17.4k2954
$endgroup$
add a comment |
$begingroup$
There is a temperature dependence, for the same molecule the lower the temperature, the smaller is the minimum $Z$. You can see this if you plot $Z$ vs $P$ using the van der Waals equation. This makes sense because at lower temperature ( and hence lower energy) molecules are more able to associate with one another and so are less 'ideal', i.e. there is less energy with which to break up intermolecular interactions.
If you plot $Z$ vs $P$ using reduced pressure, temperature and volume, as $T_r=T/T_c,; P_r=P/P_c,;V_r=V/V_c$ where the subscript $c$ indicates values at the critical point then many gases fall onto the same plots at a given temperature. This is called the 'Law of corresponding states'. The constants $a$ and $b$ are eliminated from the final equation for $Z$, although it looks as thought they will be used, making the resulting equation 'universal'. Thus all gasses behave the same way when compared at corresponding conditions'
$Z$ vs $P_r$ at different reduced temperatures
For a van der Waals gas $(P+a/V_m^2)(V_m-b)=RT$ where $a$ and $b$ are obtained from experiment for each gas and $V_m$ is the molar volume and $T_C=8a/(27Rb), ; P_c=a/(27b^2) ,; V_{mc}=3b$ where $V_{mc}$ is critical molar volume.
answered Jan 12 at 10:02
porphyrinporphyrin
17.4k2954
$endgroup$
There is a temperature dependence, for the same molecule the lower the temperature, the smaller is the minimum $Z$. You can see this if you plot $Z$ vs $P$ using the van der Waals equation. This makes sense because at lower temperature ( and hence lower energy) molecules are more able to associate with one another and so are less 'ideal', i.e. there is less energy with which to break up intermolecular interactions.
If you plot $Z$ vs $P$ using reduced pressure, temperature and volume, as $T_r=T/T_c,; P_r=P/P_c,;V_r=V/V_c$ where the subscript $c$ indicates values at the critical point then many gases fall onto the same plots at a given temperature. This is called the 'Law of corresponding states'. The constants $a$ and $b$ are eliminated from the final equation for $Z$, although it looks as thought they will be used, making the resulting equation 'universal'. Thus all gasses behave the same way when compared at corresponding conditions'
$Z$ vs $P_r$ at different reduced temperatures
For a van der Waals gas $(P+a/V_m^2)(V_m-b)=RT$ where $a$ and $b$ are obtained from experiment for each gas and $V_m$ is the molar volume and $T_C=8a/(27Rb), ; P_c=a/(27b^2) ,; V_{mc}=3b$ where $V_{mc}$ is critical molar volume.
answered Jan 12 at 10:02
porphyrinporphyrin
17.4k2954
edited Jan 12 at 12:53
answered Jan 12 at 10:02
porphyrinporphyrin
17.4k2954
answered Jan 12 at 10:02
porphyrinporphyrin
17.4k2954
answered Jan 12 at 10:02
porphyrinporphyrin
17.4k2954
17.4k2954
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add a comment |
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