Answer :
To solve for the piston height \( h \), we'll use the ideal gas law to find the volume of the compressed air, then apply Archimedes' principle to find the height at which the piston floats.
First, let's find the volume of the compressed air using the ideal gas law:
\[ PV = nRT \]
Where:
- \( P = 1 \) atmosphere (pressure)
- \( V \) = volume of the compressed air (to be determined)
- \( n = 0.11 \) mol (number of moles of air)
- \( R = 0.0821 \) atm·L/mol·K (ideal gas constant)
- \( T = 30 + 273.15 = 303.15 \) K (temperature in Kelvin)
\[ V = \frac{{nRT}}{P} \]
\[ V = \frac{{0.11 \, \text{{mol}} \times 0.0821 \, \text{{atm}} \cdot \text{{L/mol}} \cdot \text{{K}} \times 303.15 \, \text{{K}}}}{1 \, \text{{atm}}} \]
\[ V ≈ 2.76 \, \text{{L}} \]
Now, we'll find the height at which the piston floats using Archimedes' principle, which states that the buoyant force on the piston is equal to the weight of the displaced air:
\[ F_{\text{{buoyant}}} = mg \]
Where:
- \( F_{\text{{buoyant}}} \) = buoyant force
- \( m = 50 \, \text{{kg}} \) (mass of the piston)
- \( g = 9.81 \, \text{{m/s}}^2 \) (acceleration due to gravity)
The buoyant force can also be expressed as:
\[ F_{\text{{buoyant}}} = \rho_{\text{{air}}} \cdot g \cdot V_{\text{{displaced}}} \]
Where:
- \( \rho_{\text{{air}}} \) = density of air at 30°C and 1 atm pressure
- \( V_{\text{{displaced}}} \) = volume of air displaced by the piston
Since the piston is circular and the cylinder has a diameter of 10 cm, the area of the piston is \( A = \frac{{\pi \cdot d^2}}{4} = \frac{{\pi \cdot (0.1 \, \text{{m}})^2}}{4} \). The height \( h \) is then \( V_{\text{{displaced}}} = A \cdot h \).
\[ \rho_{\text{{air}}} \cdot g \cdot A \cdot h = mg \]
\[ \rho_{\text{{air}}} \cdot A \cdot h = m \]
\[ h = \frac{m}{\rho_{\text{{air}}} \cdot A} \]
To find \( \rho_{\text{{air}}} \) at 30°C and 1 atm pressure, we can use the ideal gas law:
\[ PV = nRT \]
\[ \frac{m}{\rho_{\text{{air}}}} = \frac{nRT}{P} \]
\[ \rho_{\text{{air}}} = \frac{mP}{nRT} \]
Substitute the known values:
\[ \rho_{\text{{air}}} = \frac{50 \, \text{{kg}} \times 1 \, \text{{atm}}}{0.11 \, \text{{mol}} \times 0.0821 \, \text{{atm}} \cdot \text{{L/mol}} \cdot \text{{K}} \times 303.15 \, \text{{K}}} \]
\[ \rho_{\text{{air}}} ≈ 1.62 \, \text{{kg/m}}^3 \]
Now, plug in the values to find \( h \):
\[ h = \frac{50 \, \text{{kg}}}{1.62 \, \text{{kg/m}}^3 \times \left( \frac{\pi \cdot (0.1 \, \text{{m}})^2}{4} \right)} \]
\[ h ≈ 9.74 \, \text{{m}} \]
So, the height \( h \) at which the piston floats is approximately 9.74 meters.
Hope this helps!
First, let's find the volume of the compressed air using the ideal gas law:
\[ PV = nRT \]
Where:
- \( P = 1 \) atmosphere (pressure)
- \( V \) = volume of the compressed air (to be determined)
- \( n = 0.11 \) mol (number of moles of air)
- \( R = 0.0821 \) atm·L/mol·K (ideal gas constant)
- \( T = 30 + 273.15 = 303.15 \) K (temperature in Kelvin)
\[ V = \frac{{nRT}}{P} \]
\[ V = \frac{{0.11 \, \text{{mol}} \times 0.0821 \, \text{{atm}} \cdot \text{{L/mol}} \cdot \text{{K}} \times 303.15 \, \text{{K}}}}{1 \, \text{{atm}}} \]
\[ V ≈ 2.76 \, \text{{L}} \]
Now, we'll find the height at which the piston floats using Archimedes' principle, which states that the buoyant force on the piston is equal to the weight of the displaced air:
\[ F_{\text{{buoyant}}} = mg \]
Where:
- \( F_{\text{{buoyant}}} \) = buoyant force
- \( m = 50 \, \text{{kg}} \) (mass of the piston)
- \( g = 9.81 \, \text{{m/s}}^2 \) (acceleration due to gravity)
The buoyant force can also be expressed as:
\[ F_{\text{{buoyant}}} = \rho_{\text{{air}}} \cdot g \cdot V_{\text{{displaced}}} \]
Where:
- \( \rho_{\text{{air}}} \) = density of air at 30°C and 1 atm pressure
- \( V_{\text{{displaced}}} \) = volume of air displaced by the piston
Since the piston is circular and the cylinder has a diameter of 10 cm, the area of the piston is \( A = \frac{{\pi \cdot d^2}}{4} = \frac{{\pi \cdot (0.1 \, \text{{m}})^2}}{4} \). The height \( h \) is then \( V_{\text{{displaced}}} = A \cdot h \).
\[ \rho_{\text{{air}}} \cdot g \cdot A \cdot h = mg \]
\[ \rho_{\text{{air}}} \cdot A \cdot h = m \]
\[ h = \frac{m}{\rho_{\text{{air}}} \cdot A} \]
To find \( \rho_{\text{{air}}} \) at 30°C and 1 atm pressure, we can use the ideal gas law:
\[ PV = nRT \]
\[ \frac{m}{\rho_{\text{{air}}}} = \frac{nRT}{P} \]
\[ \rho_{\text{{air}}} = \frac{mP}{nRT} \]
Substitute the known values:
\[ \rho_{\text{{air}}} = \frac{50 \, \text{{kg}} \times 1 \, \text{{atm}}}{0.11 \, \text{{mol}} \times 0.0821 \, \text{{atm}} \cdot \text{{L/mol}} \cdot \text{{K}} \times 303.15 \, \text{{K}}} \]
\[ \rho_{\text{{air}}} ≈ 1.62 \, \text{{kg/m}}^3 \]
Now, plug in the values to find \( h \):
\[ h = \frac{50 \, \text{{kg}}}{1.62 \, \text{{kg/m}}^3 \times \left( \frac{\pi \cdot (0.1 \, \text{{m}})^2}{4} \right)} \]
\[ h ≈ 9.74 \, \text{{m}} \]
So, the height \( h \) at which the piston floats is approximately 9.74 meters.
Hope this helps!
