Answer:
The substance with a weaker molecular attraction
Answer:
1.5 × 10² mL
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 1.9 atm
- Initial volume of the gas (V₁): 80 mL
- Final pressure of the gas (P₂): 1.0 atm (standard pressure)
- Final volume of the gas (V₂): ?
Step 2: Calculate the final volume of the gas
For an ideal gas, we can calculate the final volume of the gas using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 1.9 atm × 80 mL/1.0 atm
V₂ = 1.5 × 10² mL
Since the pressure decreased, the volume of the gas increased.
Answer:
The work done and heat absorbed are both -8,1 kJ
Explanation:
The work done in an isobaric process is defined as:
W = -P (Vf - Vi)
Where P is pressure ( 10 atm)
Vf = 10 L
Vi = 2 L
Thus, <em>W = -80 atm×L ≡ -8,1 kJ</em>
This is the work done in expansion of the gas. As the gas remains at the same temperature, there is no change in internal energy doing that all work was absorbed as heat.
I hope it helps!
Answer:
Region B, because the pressure inside the cylinder is equal to the vapor pressure of water at 80∘C when both liquid and gas phases are present.
Explanation:
As expansion occurs, liquid water evaporates reversibly, holding the pressure constant at the equilibrium vapor pressure of water at 80∘C(0.47atm) 80∘C (0.47 atm). When all of the liquid has evaporated, the pressure drops and follows the ideal gas law.
