A spherical balloon initially contains 25m3 of helium gas at 20o C and 150 kPa. A valve is now opened and the helium is allowed
to escape slowly. The valve is closed when the pressure inside the balloon drops to the atmospheric pressure of 100 kPa. The elasticity of the balloon material is such that the pressure inside the balloon during the process varies with the volume according to the relation P= A + BV where P is the pressure in kPa, V is the total volume in m3 , the constant A has a value of -100 kPa (i.e. negative 100kPa), and the constant B has an unknown value. The temperature at which the helium exits the balloon is constant at 15o C throughout the whole process.
Disregarding any heat transfer, determine
(a) the final temperature of helium in the balloon after the valve has been closed, and
(b) the mass of helium that has escaped. Consider helium to be an ideal gas with constant specific heats [5.193kJ/(kg K) for Cp,o and 3.116kJ/(kg K) for Cv,o].
Disregarding any heat transfer, determine
(a) the final temperature of helium in the balloon after the valve has been closed, and
(b) the mass of helium that has escaped. Consider helium to be an ideal gas with constant specific heats [5.193kJ/(kg K) for Cp,o and 3.116kJ/(kg K) for Cv,o].
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Answer:
The equation is already balanced. There's an equal number of materials on each side of the equation.
<u><em>1.5 grams of glucose is produced from 2.20 g of CO₂.</em></u>
To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.
The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.
So, in this case, the balanced reaction is:
6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:
- CO₂: 6 moles
- H₂O: 6 moles
- C₆H₁₂O₆: 1 mole
- O₂: 6 moles
So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:
moles of CO₂= 0.05 moles
Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?
moles of C₆H₁₂O₆= 8.33*10⁻³
Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:
<em>mass of glucose= 1.5 grams</em>
Then, <u><em>1.5 grams of glucose is produced from 2.20 g of CO₂.</em></u>