Answer:
The new volume is 7,606.96 Liter.
Explanation:
The combined gas equation is,
where,
= initial pressure of gas in the balloon = 0.918 atm
= final pressure of gas in the balloon = 0.0012 atm
= initial volume of gas in the balloon =
= final volume of gas in the balloon = ?
= initial temperature of gas in the balloon =
= final temperature of gas in the balloon =
Now put all the given values in the above equation, we get:
The new volume is 7,606.96 Liter.
Taking into accoun the STP conditions and the ideal gas law, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
First of all, the STP conditions refer to the standard temperature and pressure, where the values used are: pressure at 1 atmosphere and temperature at 0°C. These values are reference values for gases.
On the other side, the pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where:
Then, in this case:
Replacing in the expression for the ideal gas law:
1 atm× 44.1 L= n× 0.082 × 273 K
Solving:
n=1.97 moles
Being the molar mass of O₂, that is, the mass of one mole of the compound, 32 g/mole, the amount of mass that 1.97 moles contains can be calculated as:
= 63.04 g ≈ <u><em>63 g</em></u>
Finally, the correct answer is option e. 63 grams of O₂ are present in 44.1 L of O2 at STP.
Learn more about the ideal gas law: