The density of mercury is 1.36 X 10^4 Kg/m^3 at 0 degree Celsius. Calculate its value at 100 degree Celsius and at 22 degree Celsius. Take cubic expansivity of mercury as equal to 180 X 10^-6 K.
densityTf=1.36e4kg/m^3 / (1+Tf*1.80e-6)
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The density of mercury is 1•36×10^4kgm^3 at 0°c calculate it values at 100°c and at 22°c take 3 linear expansivity if mercury 180×10^6k^1
To calculate the density of mercury at different temperatures, we can use the formula for volumetric expansion due to temperature change:
ΔV = V₀ × β × ΔT
Where:
ΔV is the change in volume
V₀ is the initial volume
β is the coefficient of cubic expansion
ΔT is the change in temperature
Given:
Density of mercury at 0 degrees Celsius (ρ₀) = 1.36 × 10^4 kg/m³
Cubic expansivity of mercury (β) = 180 × 10^-6 K
Now, let's calculate the density of mercury at 100 degrees Celsius (ρ₁):
First, we need to calculate the change in volume (ΔV) from 0 to 100 degrees Celsius:
ΔV = V₁ - V₀
We know that density (ρ) is equal to mass (m) divided by volume (V):
ρ = m/V
Rearranging the equation to solve for volume:
V = m/ρ
So, ΔV = Δm / ρ₀
Where Δm is the change in mass, which is equal to 0 since we are considering the same amount of mercury. Therefore, ΔV = 0.
Now, let's calculate the density at 100 degrees Celsius:
ρ₁ = ρ₀ + Δρ
= ρ₀ + (Δm / V₁)
= ρ₀ + 0
= ρ₀
Thus, the density of mercury at 0 degrees Celsius is the same as the density at 100 degrees Celsius.
Now, let's calculate the density of mercury at 22 degrees Celsius (ρ₂):
Again, we need to calculate the change in volume (ΔV) from 0 to 22 degrees Celsius. Using the same formula as above:
ΔV = V₂ - V₀
To calculate ΔV, we need to consider the change in temperature (ΔT):
ΔV = V₀ × β × ΔT
Substituting the values we have:
ΔV = V₀ × β × ΔT
= V₀ × 180 × 10^-6 × 22
Since ρ = m/V, Δm = ρ₀ × ΔV
Therefore, the change in mass (Δm) is:
Δm = ρ₀ × ΔV
= ρ₀ × (V₀ × 180 × 10^-6 × 22)
Now, let's calculate the density at 22 degrees Celsius:
ρ₂ = ρ₀ + (Δm / V₂)
= ρ₀ + [(ρ₀ × (V₀ × 180 × 10^-6 × 22)) / V₂]
We already know the value of ρ₀, and V₀ is the reciprocal of density (since Density = mass/volume). So, V₀ = 1/ρ₀.
Substituting these values:
ρ₂ = ρ₀ + [(ρ₀ × (1/ρ₀) × 180 × 10^-6 × 22) / V₂]
= ρ₀ + (180 × 10^-6 × 22) / V₂
To calculate V₂, we can use the formula V = m/ρ. Since we know the density at 22 degrees Celsius (ρ₂), we can substitute this value for ρ and solve for V:
V₂ = m/ρ₂
Substituting this value back into the density equation, we get:
ρ₂ = ρ₀ + (180 × 10^-6 × 22) / (m/ρ₂)
= ρ₀ + [(180 × 10^-6 × 22) × ρ₂] / m
Now, using the equation V₀ = 1/ρ₀, we can substitute this value into the equation:
ρ₂ = (1/V₀) + [(180 × 10^-6 × 22) × ρ₂] / m
Since the mass (m) does not change, we can move it outside the equation:
ρ₂ = (1/V₀) + [(180 × 10^-6 × 22) × ρ₂] × (1/m)
Now, let's rearrange the equation to solve for ρ₂:
ρ₂ - [(180 × 10^-6 × 22) × ρ₂] × (1/m) = 1/V₀
ρ₂ - [(180 × 10^-6 × 22) / m] × ρ₂ = 1/V₀
ρ₂ × (1 - [(180 × 10^-6 × 22) / m]) = 1/V₀
ρ₂ = 1 / [V₀ × (1 - [(180 × 10^-6 × 22) / m])]
Finally, substitute the value of 1/V₀, which is ρ₀, into the equation:
ρ₂ = ρ₀ / (1 - [(180 × 10^-6 × 22) / m])
Note that you have not provided the mass of the mercury. Without it, we cannot calculate the density of mercury at 22 degrees Celsius.