Answer:
a) 3.39 × 10²³ atoms
b) 6.04 × 10⁻²¹ J
c) 1349.35 m/s
Explanation:
Given:
Diameter of the balloon, d = 29.6 cm = 0.296 m
Temperature, T = 19.0° C = 19 + 273 = 292 K
Pressure, P = 1.00 atm = 1.013 × 10⁵ Pa
Volume of the balloon =
or
Volume of the balloon =
or
Volume of the balloon, V = 0.0135 m³
Now,
From the relation,
PV = nRT
where,
n is the number of moles
R is the ideal gas constant = 8.314 kg⋅m²/s²⋅K⋅mol
on substituting the respective values, we get
1.013 × 10⁵ × 0.0135 = n × 8.314 × 292
or
n = 0.563
1 mol = 6.022 × 10²³ atoms
Thus,
0.563 moles will have = 0.563 × 6.022 × 10²³ atoms = 3.39 × 10²³ atoms
b) Average kinetic energy =
where,
Boltzmann constant,
Average kinetic energy =
or
Average kinetic energy = 6.04 × 10⁻²¹ J
c) rms speed =
where, m is the molar mass of the Helium = 0.004 Kg
or
rms speed =
or
rms speed = 1349.35 m/s
Answer:
The speed is
Explanation:
From the question w are told that
The angle made is
The distance above the surface of the water is
The value of
The maximum height attained by the fish is mathematically evaluate as
Making v which is the speed of the fish the subject of the formula
substituting values
Answer:
The answers to the question are
(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W
(ii) The temperature of the outer surface of the insulation is 49.89 °C
Explanation:
To solve the question, we note that the heat transferred is given by
Where
= Temperature at the inside of the pipe = 300 °C
= Temperature at the outside of the pipe = 20 °C
r₁ =internal radius of pipe = 4.0 cm
r₂ = Outer radius of pipe = 4.5 cm
r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm
= 15 W/m·K
= 0.038 W/m·K
= 75 W/m²·K
= 10 W/m²·K
Plugging in the values in the above equation where for a unit length L = 1 m, we have
Q = 131.32 W
From which we have, for the film of air at the pipe outer boundary layer
Where
for the air film on the pipe outer surface is given by
where A =area of the outside of the pipe
= = 0.227 K/W
Therefore
131.32 W = which gives
= 49.89 °C
Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)
Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)
T₁ = Surface temperature of the pipe = 49.89 °C and
T₂ = Temperature of the surrounding = 20.00 °C
Plugging in the values gives, q' = 0.307 W per m²
Total heat lost per unit length = 131.32 + 0.307 =131.62 W