Mechanical twinning occurs in metals having which type(s) of crystal structure(s)?
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The Boyle-Mariotte's law or Boyle's law is one of the laws of gases that <u>relates the volume (V) and pressure (P) of a certain amount of gas maintained at constant temperature</u>, as follows:
PV = k
where k is a constant.
We can relate the state of a gas at a specific pressure and volume to another state in which the same gas is at different P and V since the product of both variables is equal to a constant, according to the Boyle's law, which will be the same regardless of the state of the gas. In this way,
P₁V₁ = P₂V₂
Where P₁ and V₁ is the pressure and volume of the gas to a state 1 and P₂ and V₂ is the pressure and volume of the same gas in a state 2.
In this case, in the state 1 the gas occupies a volume V₁ = 100 mL at a pressure of P₁ = 150 kPa. Then, in the state 2 the gas occupies a volume V₂ (that we must calculate through the boyle's law) at a pressure of P₂ = 200 kPa. Substituting these values in the previous equation and clearing V₂, we have,
P₁V₁ = P₂V₂ → V₂ =
→ V₂ =
→ V₂ = 75 mL
Then, the volume occupied by the gas at 200 kPa is V₂ = 75 mL
Answer: The entropy change of the surroundings will be -17.7 J/K mol.
Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol
Amount of Acetone given = 10.8 g
Number of moles is calculated by using the formula:
Molar mass of acetone = 58 g/mol
Number of moles =
If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then
0.1862 moles will have =
To calculate the entropy change for the system, we use the formula:
Temperature = 56.2°C = (273 + 56.2)K = 329.2K
Putting values in above equation, we get
(Conversion Factor: 1 kJ = 1000J)
At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,