QUESTION IMAGE
Question
- complete the following table for an ideal gas. p (atm) v (l) n (mol) t a. 5.00 2.00 155°c b. 0.300 2.00 155 k c. 4.47 25.0 2.01 d. 2.25 10.5 75°c
Response
- Recall the ideal - gas law:
- The ideal - gas law is given by \(PV = nRT\), where \(P\) is the pressure in atmospheres (atm), \(V\) is the volume in liters (L), \(n\) is the number of moles, \(T\) is the temperature in Kelvin (K), and \(R=0.08206\ L\cdot atm/(mol\cdot K)\).
- a. Calculate the volume (\(V\)):
- First, convert the temperature to Kelvin. \(T=(155 + 273.15)\ K=428.15\ K\).
- From the ideal - gas law \(V=\frac{nRT}{P}\).
- Substitute \(n = 2.00\ mol\), \(R = 0.08206\ L\cdot atm/(mol\cdot K)\), \(T = 428.15\ K\), and \(P = 5.00\ atm\) into the formula.
- \(V=\frac{2.00\ mol\times0.08206\ L\cdot atm/(mol\cdot K)\times428.15\ K}{5.00\ atm}\)
- \(V=\frac{2.00\times0.08206\times428.15}{5.00}\ L\)
- \(V=\frac{70.27}{5.00}\ L = 14.1\ L\).
- b. Calculate the number of moles (\(n\)):
- Use the ideal - gas law \(n=\frac{PV}{RT}\).
- Substitute \(P = 0.300\ atm\), \(V = 2.00\ L\), \(R = 0.08206\ L\cdot atm/(mol\cdot K)\), and \(T = 155\ K\) into the formula.
- \(n=\frac{0.300\ atm\times2.00\ L}{0.08206\ L\cdot atm/(mol\cdot K)\times155\ K}\)
- \(n=\frac{0.600}{12.72}\ mol=0.0472\ mol\).
- c. Calculate the temperature (\(T\)):
- From the ideal - gas law \(T=\frac{PV}{nR}\).
- Substitute \(P = 4.47\ atm\), \(V = 25.0\ L\), \(n = 2.01\ mol\), and \(R = 0.08206\ L\cdot atm/(mol\cdot K)\) into the formula.
- \(T=\frac{4.47\ atm\times25.0\ L}{2.01\ mol\times0.08206\ L\cdot atm/(mol\cdot K)}\)
- \(T=\frac{111.75}{0.16494\ mol\cdot K}\)
- \(T = 677\ K\).
- d. Calculate the pressure (\(P\)):
- First, convert the temperature to Kelvin. \(T=(75 + 273.15)\ K = 348.15\ K\).
- From the ideal - gas law \(P=\frac{nRT}{V}\).
- Substitute \(n = 10.5\ mol\), \(R = 0.08206\ L\cdot atm/(mol\cdot K)\), \(T = 348.15\ K\), and \(V = 2.25\ L\) into the formula.
- \(P=\frac{10.5\ mol\times0.08206\ L\cdot atm/(mol\cdot K)\times348.15\ K}{2.25\ L}\)
- \(P=\frac{10.5\times0.08206\times348.15}{2.25}\ atm\)
- \(P=\frac{301.7}{2.25}\ atm = 134\ atm\).
The completed table is:
| \(P\) (atm) | \(V\) (L) | \(n\) (mol) | \(T\) (K) |
|---|---|---|---|
| \(0.300\) | \(2.00\) | \(0.0472\) | \(155\) |
| \(4.47\) | \(25.0\) | \(2.01\) | \(677\) |
| \(134\) | \(2.25\) | \(10.5\) | \(348.15\) |
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- Recall the ideal - gas law:
- The ideal - gas law is given by \(PV = nRT\), where \(P\) is the pressure in atmospheres (atm), \(V\) is the volume in liters (L), \(n\) is the number of moles, \(T\) is the temperature in Kelvin (K), and \(R=0.08206\ L\cdot atm/(mol\cdot K)\).
- a. Calculate the volume (\(V\)):
- First, convert the temperature to Kelvin. \(T=(155 + 273.15)\ K=428.15\ K\).
- From the ideal - gas law \(V=\frac{nRT}{P}\).
- Substitute \(n = 2.00\ mol\), \(R = 0.08206\ L\cdot atm/(mol\cdot K)\), \(T = 428.15\ K\), and \(P = 5.00\ atm\) into the formula.
- \(V=\frac{2.00\ mol\times0.08206\ L\cdot atm/(mol\cdot K)\times428.15\ K}{5.00\ atm}\)
- \(V=\frac{2.00\times0.08206\times428.15}{5.00}\ L\)
- \(V=\frac{70.27}{5.00}\ L = 14.1\ L\).
- b. Calculate the number of moles (\(n\)):
- Use the ideal - gas law \(n=\frac{PV}{RT}\).
- Substitute \(P = 0.300\ atm\), \(V = 2.00\ L\), \(R = 0.08206\ L\cdot atm/(mol\cdot K)\), and \(T = 155\ K\) into the formula.
- \(n=\frac{0.300\ atm\times2.00\ L}{0.08206\ L\cdot atm/(mol\cdot K)\times155\ K}\)
- \(n=\frac{0.600}{12.72}\ mol=0.0472\ mol\).
- c. Calculate the temperature (\(T\)):
- From the ideal - gas law \(T=\frac{PV}{nR}\).
- Substitute \(P = 4.47\ atm\), \(V = 25.0\ L\), \(n = 2.01\ mol\), and \(R = 0.08206\ L\cdot atm/(mol\cdot K)\) into the formula.
- \(T=\frac{4.47\ atm\times25.0\ L}{2.01\ mol\times0.08206\ L\cdot atm/(mol\cdot K)}\)
- \(T=\frac{111.75}{0.16494\ mol\cdot K}\)
- \(T = 677\ K\).
- d. Calculate the pressure (\(P\)):
- First, convert the temperature to Kelvin. \(T=(75 + 273.15)\ K = 348.15\ K\).
- From the ideal - gas law \(P=\frac{nRT}{V}\).
- Substitute \(n = 10.5\ mol\), \(R = 0.08206\ L\cdot atm/(mol\cdot K)\), \(T = 348.15\ K\), and \(V = 2.25\ L\) into the formula.
- \(P=\frac{10.5\ mol\times0.08206\ L\cdot atm/(mol\cdot K)\times348.15\ K}{2.25\ L}\)
- \(P=\frac{10.5\times0.08206\times348.15}{2.25}\ atm\)
- \(P=\frac{301.7}{2.25}\ atm = 134\ atm\).
The completed table is:
| \(P\) (atm) | \(V\) (L) | \(n\) (mol) | \(T\) (K) |
|---|---|---|---|
| \(0.300\) | \(2.00\) | \(0.0472\) | \(155\) |
| \(4.47\) | \(25.0\) | \(2.01\) | \(677\) |
| \(134\) | \(2.25\) | \(10.5\) | \(348.15\) |