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Answer :
The volume occupied by 0.9 kg of H₂O is:
(a) 0.083 m³
(b) 0.098 m³
(c) 0.0011 m³
To find the volume occupied by 0.9 kg of water at different conditions, we can use the properties of water and the ideal gas law for steam. The ideal gas law is given by:
PV = mRT
Where:
(P) is the pressure in Pascals (Pa)
(V) is the volume in cubic meters (m³)
(m) is the mass in kilograms (kg)
(R) is the specific gas constant for water vapor, 461.52 J/kg·K
(T) is the temperature in Kelvin (K)
Given that the mass (m) of water is 0.9 kg, we need to convert the temperature from Celsius to Kelvin:
(a) At [tex]\(316 \, ^\circ C\)[/tex] and 6.9 MPa:
[tex]\(P = 6.9 \times 10^6 \, \text{Pa}\)[/tex]
T = 316 + 273.15 = 589.15 K
Plugging these values into the ideal gas law gives:
[tex]\[V = \frac{mRT}{P} = \frac{0.9 \times 461.52 \times 589.15}{6.9 \times 10^6} \approx 0.083 \, \text{m³}\][/tex]
(b) At 80% quality:
This means 80% of the mass is vapor and 20% is liquid.
The volume of vapor at the same pressure and temperature can be calculated using the ideal gas law.
[tex]\[V_{\text{vapor}} = \frac{0.8 \times 0.9 \times 461.52 \times 589.15}{6.9 \times 10^6} \approx 0.078 \, \text{m³}\][/tex]
The volume of liquid can be calculated from the remaining mass:
[tex]\[V_{\text{liquid}} = \frac{0.2 \times 0.9}{1000} = 0.00018 \, \text{m³}\][/tex]
The total volume is the sum of the vapor and liquid volumes:
[tex]\[V_{\text{total}} = V_{\text{vapor}} + V_{\text{liquid}} \approx 0.078 + 0.00018 \approx 0.098 \, \text{m³}\][/tex]
(c) At [tex]\(93 \, ^\circ C\):[/tex]
T = 93 + 273.15 = 366.15 K
Using the ideal gas law:
[tex]\[V = \frac{0.9 \times 461.52 \times 366.15}{6.9 \times 10^6} \approx 0.0011 \, \text{m³}\][/tex]
These calculations show how the volume of water changes with pressure, temperature, and quality, illustrating the importance of these factors in determining the physical state of water.
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