Answer:
Explanation:
1)
Given data:
Initial volume of balloon = 0.8 L
Initial temperature = 12°C ( 12+273= 285 K)
Final temperature = 300°C (300+273 = 573 K)
Final volume = ?
Solution:
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 0.8 L .573 K / 285 K
V₂ = 458.4 L / 285
V₂ = 1.61 L
2)
Initial pressure = 204 kpa
Initial temperature = 29°C ( 29 + 273 = 302 K)
Final temperature = ?
Final pressure = 300 kpa
Solution:
P₁/T₁ = P₂/T₂
T₂ = T₁P₂/P₁
T₂ = 302 K . 300 kpa / 204 kpa
T₂ = 90600 K/ 204
T₂ = 444.12 K
3)
Given data:
Initial volume = 14 L
Initial pressure = 2.1 atm
Initial temperature = 100 K
Final temperature = 450 K
Final volume = ?
Final pressure = 1.2 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2.1 atm × 14 L × 450 K / 100 K × 1.2 atm
V₂ = 13230 L / 120
V₂ = 110.25 L
Answer:
About 0.1738 liters
Explanation:
Using the formula PV=nRT, where p represents pressure in atmospheres, v represents volume in liters, n represents the number of moles of ideal gas, R represents the ideal gas constant, and T represents the temperature in kelvin, you can solve this problem. But first, you need to convert to the proper units. 215ml=0.215L, 86.4kPa is about 0.8527 atmospheres, and 15C is 288K. Plugging this into the equation, you get:

Now that you know the number of moles of gas, you can plug back into the equation with STP conditions:

Hope this helps!
1 mole ----------- 22.4 L ( at STP )
3.5 mols --------- ?
V = 3.5 x 22.4 / 1
V = 78.4 L
hope this helps!
Answer:
Part A
The volume of the gaseous product is
Part B
The volume of the the engine’s gaseous exhaust is
Explanation:
Part A
From the question we are told that
The temperature is 
The pressure is 
The of 
The chemical equation for this combustion is

The number of moles of
that reacted is mathematically represented as

The molar mass of
is constant value which is
So 

The gaseous product in the reaction is
and water vapour
Now from the reaction
2 moles of
will react with 25 moles of
to give (16 + 18) moles of
and 
So
1 mole of
will react with 12.5 moles of
to give 17 moles of
and 
This implies that
0.8754 moles of
will react with (12.5 * 0.8754 ) moles of
to give (17 * 0.8754) of
and 
So the no of moles of gaseous product is


From the ideal gas law

making V the subject

Where R is the gas constant with a value 
Substituting values
Part B
From the reaction the number of moles of oxygen that reacted is


The volume is


No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as

Substituting values