Answer:
19063.6051 g
Explanation:
Pressure = Atmospheric pressure + Gauge Pressure
Atmospheric pressure = 97 kPa
Gauge pressure = 500 kPa
Total pressure = 500 + 97 kPa = 597 kPa
Also, P (kPa) = 1/101.325 P(atm)
Pressure = 5.89193 atm
Volume = 2.5 m³ = 2500 L ( As m³ = 1000 L)
Temperature = 28 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (28.2 + 273.15) K = 301.15 K
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
5.89193 atm × 2500 L = n × 0.0821 L.atm/K.mol × 301.15 K
⇒n = 595.76 moles
Molar mass of oxygen gas = 31.9988 g/mol
Mass = Moles * Molar mass = 595.76 * 31.9988 g = 19063.6051 g
Answer:
A transformer is an electric device that uses electromagnetism to change voltage from one level to another or to isolate one voltage from another.
Answer:
(a) Increases
(b) Increases
(c) Increases
(d) Increases
(e) Decreases
Explanation:
The tensile modulus of a semi-crystalline polymer depends on the given factors as:
(a) Molecular Weight:
It increases with the increase in the molecular weight of the polymer.
(b) Degree of crystallinity:
Tensile strength of the semi-crystalline polymer increases with the increase in the degree of crystallinity of the polymer.
(c) Deformation by drawing:
The deformation by drawing in the polymer results in the finely oriented chain structure of the polymer with the greater inter chain secondary bonding structure resulting in the increase in the tensile strength of the polymer.
(d) Annealing of an undeformed material:
This also results in an increase in the tensile strength of the material.
(e) Annealing of a drawn material:
A semi crystalline material which is drawn when annealed results in the decreased tensile strength of the material.
Answer:
c from transmitter to a receiver
Answer:
Uair = 0.0749 KW/k = 74.9 W/k
Explanation:
The natural air change per hour is given by the formula:
Natural Air Change per Hour = ACPH = 60*Volume Flow/Volume
where,
ACPH = 0.4
Volume Flow = ? in ft³/min
Volume = 19456 ft³
Therefore,
0.4 = (60 min)(Volume Flow)/(19456 ft³)
Volume Flow = (0.4)(19456 ft³)/(60 min) = (129.7 ft³/min)(1 min/60 s)
Volume Flow = (2.16 ft³/s)(0.3048 m/1 ft)³ = 0.061 m³/s
Now, we find heat loss coefficient:
Uair = Volumetric Flow*Density of air*Specific Heat Capacity of air
Uair = (0.061 m³/s)(1.225 kg/m³)(1 KJ/kg.k)
<u>Uair = 0.0749 KW/k = 74.9 W/k</u>
