the greatest amount of work is required if the process is adiabatic.The correct option is adiabatic.
The process in which heat is constant is called adiabatic process.
The The process in which temperature is constant is called isothermal process.
The process in which pressure is constant is called isobaric process.
The P-V diagram for adiabatic , isothermal and isobaric process is given below.
Work done in process = area encloses by P-V diagram axis . Since area under the curve is maximum for adiabatic process which is shown in the above diagram. So, work done by the gas will be maximum for adiabatic process.
learn more about adiabatic process.
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Answer:
86.3 g of N₂ are in the room
Explanation:
First of all we need the pressure from the N₂ in order to apply the Ideal Gases Law and determine, the moles of gas that are contained in the room.
We apply the mole fraction:
Mole fraction N₂ = N₂ pressure / Total pressure
0.78 . 1 atm = 0.78 atm → N₂ pressure
Room temperature → 20°C → 20°C + 273 = 293K
Let's replace data: 0.78 atm . 95L = n . 0.082 . 293K
(0.78 atm . 95L) /0.082 . 293K = n
3.08 moles = n
Let's convert the moles to mass → 3.08 mol . 28g /1mol = 86.3 g
Answer:
2.48626 x 10^24
Explanation:
We multiple 4.13 by avogadro's number to get that.
This is an exercise in the general or combined gas law.
To start solving this exercise, we obtain the following data:
<h3>
Data:</h3>
- T₁ = 22.5 °C + 273 = 295.5 K
- P₁ = 1.95 atm
- V₁ = ¿?
- P₂ = 3.69 atm
- T₂ = 11.9 °C + 273 = 284.9 k
- V₂= 56.4 ml
We use the following formula:
P₁V₁T₂ = P₂V₂T₁ ⇒ General formula
Where
- P₁ = Initial pressure
- V₁ = Initial volume
- T₂ = Initial temperature
- P₂ = Final pressure
- V₂ = final volume
- T₁ = Initial temperature
We clear the formula for the initial volume:
We substitute our data into the formula to solve:
The helium-filled balloon has a volume of <u>110.697 ml.</u>
