According to KE = (3/2)kT
reducing temperature, in KELVIN, by half, average KE is reduced by half.
Answer:
a.) volume = 70cm^3
b.) mass = 805 grams
Explanation:
see the explanation in attachment.
Given:
ρ = 13.6 x 10³ kg/m³, density of mercury
W = 6.0 N, weight of the mercury sample
g = 9.81 m/s², acceleration due to gravity.
Let V = the volume of the sample.
Then
W = ρVg
or
V = W/(ρg)
= (6.0 N)/[(13.6 x 10³ kg/m³)*(9.81 m/s²)]
= 4.4972 x 10⁻⁵ m³
Answer: The volume is 44.972 x 10⁻⁶ m³
We Know, F = m.a
Here, m = 2 Kg
& a = 2 m/s²
Substitute for it,
F = 2*2 Kgm/s²
F = 4 Kgm/s²
Answer:
The change in the internal energy of the gas 1,595 J
Explanation:
The first law of thermodynamics establishes that in an isolated system energy is neither created nor destroyed, but undergoes transformations; If mechanical work is applied to a system, its internal energy varies; If the system is not isolated, part of the energy is transformed into heat that can leave or enter the system; and finally an isolated system is an adiabatic system (heat can neither enter nor exit, so no heat transfer takes place.)
This is summarized in the expression:
ΔU= Q - W
where the heat absorbed and the work done by the system on the environment are considered positive.
Taking these considerations into account, in this case:
- Q= 500 cal= 2,092 J (being 1 cal=4.184 J)
Replacing:
ΔU= 2,092 J - 500 J
ΔU= 1,592 J whose closest answer is 1,595 J
<u><em>The change in the internal energy of the gas 1,595 J
</em></u>
