Answer:
V₂ =31.8 mL
Explanation:
Given data:
Initial volume of gas = 45 mL
Initial temperature = 135°C (135+273 =408 K)
Final temperature = 15°C (15+273 =288 K)
Final volume of gas = ?
Solution:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 45 mL × 288 K / 408 k
V₂ = 12960 mL.K / 408 K
V₂ =31.8 mL
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Answer:
Explanation:
We are to carefully sketch a curve that relates to the potential energy of two O atoms versus the distance between their nuclei.
From the diagram, O2 have higher potential energy than the N2 molecule. Because on the periodic table, the atomic size increases from left to right on across the period, thus O2 posses a larger atomic size than N2 atom.
Therefore, the bond length formation between the two O atoms will be larger compared to that of the two N atoms.
Answer:
1. ΔE = 0 J
2. ΔH = 0 J
3. q = 3.2 × 10³ J
4. w = -3.2 × 10³ J
Explanation:
The change in the internal energy (ΔE) and the change in the enthalpy (ΔH) are functions of the temperature. If the temperature is constant, ΔE = 0 and ΔH = 0.
The gas initially occupies a volume V₁ = 20.0 L at P₁ = 3.2 atm. When the pressure changes to P₂ = 1.6 atm, we can find the volume V₂ using Boyle's law.
P₁ × V₁ = P₂ × V₂
3.2 atm × 20.0 L = 1.6 atm × V₂
V₂ = 40 L
The work (w) can be calculated using the following expression.
w = - P . ΔV
where,
P is the external pressure for which the process happened
ΔV is the change in the volume
w = -1.6 atm × (40L - 20.0L) = -32 atm.L × (101.325 J/1atm.L) = -3.2 × 10³ J
The change in the internal energy is:
ΔE = q + w
0 = q + w
q = - w = 3.2 × 10³ J
