- Methyl alcohol has a specific gravity of 0.789. Calculate its density and its specific weight.
- A cylindrical container is 150 mm in diameter and weighs 2.25 N when empty. When filled to a depth of 200 mm with a certain oil, it weighs 35.4. Calculate the specific gravity of the oil.
- A storage vessel for gasoline (sg = 0.68) is a vertical cylinder 10 m in diameter. If it is filled to a depth of 6.75 m, calculate the weight and mass of the gasoline.
- Compute the pressure change required to cause to decrease in the volume of ethyl alcohol by 1%.
The Correct Answer and Explanation is :
Let’s address each question in turn:
5. Methyl alcohol has a specific gravity of 0.789. Calculate its density and its specific weight.
- Density: pecific gravity is the ratio of the substance’s density to the density of water. Since the density of water is approximately 1000 kg/m³, the density of methyl alcohol is:
calculatorturn0calculator0 - Specific Weight: pecific weight is the weight per unit volume, calculated as density multiplied by the acceleration due to gravity (9.81 m/s²):
calculatorturn0calculator1
6. A cylindrical container is 150 mm in diameter and weighs 2.25 N when empty. When filled to a depth of 200 mm with a certain oil, it weighs 35.4 N. Calculate the specific gravity of the oil.
- Calculate the weight of the oil: ubtract the empty weight of the container from the filled weight:
calculatorturn0calculator2 - Calculate the volume of the oil: he volume ( V ) of a cylinder is given by ( V = \pi r^2 h ), where ( r ) is the radius and ( h ) is the height. Converting dimensions to meters:
- Radius ( r = 150\, \text{mm} / 2 = 75\, \text{mm} = 0.075\, \text{m} )
- Height ( h = 200\, \text{mm} = 0.2\, \text{m} ) hus, the volume is:
[ V = \pi \times (0.075\, \text{m})^2 \times 0.2\, \text{m} \approx 0.003534\, \text{m}^3 ]
- Calculate the density of the oil: ensity ( \rho ) is mass per unit volume. First, find the mass by dividing the weight by gravity:
[ \text{Mass} = \frac{33.15\, \text{N}}{9.81\, \text{m/s}^2} \approx 3.38\, \text{kg} ]
hen, density:
[ \rho = \frac{3.38\, \text{kg}}{0.003534\, \text{m}^3} \approx 956\, \text{kg/m}^3 ] - Calculate the specific gravity: pecific gravity is the ratio of the oil’s density to the density of water:
[ \text{Specific Gravity} = \frac{956\, \text{kg/m}^3}{1000\, \text{kg/m}^3} = 0.956 ]
7. A storage vessel for gasoline (sg = 0.68) is a vertical cylinder 10 m in diameter. If it is filled to a depth of 6.75 m, calculate the weight and mass of the gasoline. - Calculate the volume of gasoline:
- Radius ( r = 10\, \text{m} / 2 = 5\, \text{m} )
- Height ( h = 6.75\, \text{m} ) [ V = \pi \times (5\, \text{m})^2 \times 6.75\, \text{m} \approx 530.14\, \text{m}^3 ]
- Calculate the density of gasoline: sing the specific gravity:
[ \rho = 0.68 \times 1000\, \text{kg/m}^3 = 680\, \text{kg/m}^3 ] - Calculate the mass of gasoline: [ \text{Mass} = \rho \times V = 680\, \text{kg/m}^3 \times 530.14\, \text{m}^3 \approx 360,495\, \text{kg} ]
- Calculate the weight of gasoline: [ \text{Weight} = \text{Mass} \times g = 360,495\, \text{kg} \times 9.81\, \text{m/s}^2 \approx 3,537,454\, \text{N} ]
8. Compute the pressure change required to cause a decrease in the volume of ethyl alcohol by 1%.
he bulk modulus ( K ) is defined as:
[ K = -V \frac{dP}{dV} ]
earranging to solve for the pressure change ( dP ):
[ dP = -K \frac{dV}{V} ]
iven:
- ( \frac{dV}{V} = -0.01 ) (a 1% decrease in volume)
- ( K ) for ethyl alcohol is approximately ( 1.06 \times 10^9\, \text{Pa} ) citeturn0search0
alculating ( dP ):
[ dP = 1.06 \times 10^9\, \text{Pa} \times 0.01 = 1.06 \times 10^7\, \text{Pa} ]
onverting to kilopascals:
[ dP = 1.06 \times 10^7\, \text{Pa} = 10,600\, \text{kPa} ]
herefore, a pressure increase of approximately 10,600 kPa is required to achieve a 1% decrease in the volume of ethyl alcohol.
Explanation:
he bulk modulus of a substance quantifies its resistance to uniform compression; it is defined