Ideal gas law/Tutorials: Difference between revisions
imported>Paul Wormer (2 extra problems) |
imported>Paul Wormer No edit summary |
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* <i> The molar gas constant</i> ''R'' = 0.082057 atm⋅L/(K⋅mol) | * <i> The molar gas constant</i> ''R'' = 0.082057 atm⋅L/(K⋅mol) | ||
* 1 bar = 0.98692 atm | <!--* 1 bar = 0.98692 atm --> | ||
== Example problems == | == Example problems == | ||
===Problem 1=== | ===Problem 1=== | ||
Determine the volume of 1 mol of ideal gas at pressure 1 atm and temperature 20 °C. | Determine the volume of 1 mol of ideal gas at pressure 1 atm and temperature 20 °C. | ||
:<math> | :<math> | ||
V = \frac{n\,R\,T}{p} = \frac{1\cdot 0.082057\cdot (20+273.15)}{1} \quad | V = \frac{n\,R\,T}{p} = \frac{1\cdot 0.082057\cdot (20+273.15)}{1} \quad\left[ | ||
\frac{ \mathrm{mol}\cdot\frac {\mathrm{atm}\cdot\mathrm{L}} {\mathrm{K}\cdot\mathrm{mol}} | \frac{ \mathrm{mol}\cdot\frac {\mathrm{atm}\cdot\mathrm{L}} {\mathrm{K}\cdot\mathrm{mol}} | ||
\cdot\mathrm{K} } | \cdot\mathrm{K} } | ||
{\mathrm{atm}} | {\mathrm{atm}} \right] | ||
= 24.0550 \quad \mathrm{L} | = 24.0550 \quad [\mathrm{L}] | ||
</math> | </math> | ||
===Problem 2=== | ===Problem 2=== | ||
Compute from Charles' and Gay-Lussac's law (V/T is constant) the volume of an ideal gas at 1 atm and 0 °C (Use the final result of the previous problem). Write ''V''<sub>''T''</sub> for the volume at ''T'' °C, then | Compute from Charles' and Gay-Lussac's law (''V''/''T'' is constant) the volume of an ideal gas at 1 atm and 0 °C (Use the final result of the previous problem). Write ''V''<sub>''T''</sub> for the volume at ''T'' °C, then | ||
:<math> | :<math> | ||
\frac{V_{20}}{273.15+20} = \frac{V_0}{273.15+0} \quad\Longrightarrow | \frac{V_{20}}{273.15+20} = \frac{V_0}{273.15+0} \quad\Longrightarrow | ||
V_0 = 273.15 \times \frac{24.0550}{298.15} = 22.4139 \;\mathrm{L} | V_0 = 273.15 \times \frac{24.0550}{298.15} = 22.4139\; \;[\mathrm{L}] | ||
</math> | </math> | ||
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Inserting the given numbers | Inserting the given numbers | ||
:<math> | :<math> | ||
(1.3)\qquad\qquad V_\mathrm{f} = \left(\frac{1\cdot 2}{5}\right)\; \frac{\mathrm{atm}\sdot\mathrm{L}}{\mathrm{atm}} = 0.4\; \mathrm{L} | (1.3)\qquad\qquad V_\mathrm{f} = \left(\frac{1\cdot 2}{5}\right)\;\left[ \frac{\mathrm{atm}\sdot\mathrm{L}}{\mathrm{atm}} \right] = 0.4\; [\mathrm{L}] | ||
</math> | </math> | ||
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:<math> | :<math> | ||
n=\frac{p\,V}{R\,T} = \frac{10.0\cdot 50.0} {0.0821 \cdot (273+25.0)} | n=\frac{p\,V}{R\,T} = \frac{10.0\cdot 50.0} {0.0821 \cdot (273+25.0)} | ||
\quad | \quad \left[ | ||
\frac{\mathrm{atm}\cdot \mathrm{L}}{\frac{\mathrm{atm} \cdot \mathrm{L}}{\mathrm{K}\cdot \mathrm{mol}}\cdot\mathrm{K}} | \frac{\mathrm{atm}\cdot \mathrm{L}}{\frac{\mathrm{atm} \cdot \mathrm{L}}{\mathrm{K}\cdot \mathrm{mol}}\cdot\mathrm{K}} \right] | ||
=\frac{500}{0.0821 \cdot 298}\quad \frac{\mathrm{mol} \cdot \mathrm{atm}\cdot \mathrm{L}}{\mathrm{atm}\cdot \mathrm{L}} = 20.4 \quad \mathrm{mol} | =\frac{500}{0.0821 \cdot 298}\quad\left[ \frac{\mathrm{mol} \cdot \mathrm{atm}\cdot \mathrm{L}}{\mathrm{atm}\cdot \mathrm{L}} \right] = 20.4 \quad [\mathrm{mol}] | ||
</math> | </math> |
Revision as of 07:37, 7 January 2009
- All gases mentioned below are assumed to be ideal, i.e. their p, V, T dependence is given by the ideal gas law.
- Absolute temperature is given by K = °C + 273.15.
- All pressures are absolute.
- The molar gas constant R = 0.082057 atm⋅L/(K⋅mol)
Example problems
Problem 1
Determine the volume of 1 mol of ideal gas at pressure 1 atm and temperature 20 °C.
Problem 2
Compute from Charles' and Gay-Lussac's law (V/T is constant) the volume of an ideal gas at 1 atm and 0 °C (Use the final result of the previous problem). Write VT for the volume at T °C, then
Problem 3
A certain amount of gas that has an initial pressure of 1 atm and an initial volume of 2 L, is compressed to a final pressure of 5 atm at constant temperature. What is the final volume of the gas?
Boyle's law (pV is constant)
or
Inserting the given numbers
Ideal gas law
The number n of moles is constant
It is given that the initial and final temperature are equal, , therefore the products RT on both sides of the equation cancel, and Eq. (1.4) reduces to Eq. (1.1).
Problem 4
How many moles of nitrogen are present in a 50 L tank at 25 °C when the pressure is 10 atm? Numbers include only 3 significant figures.