Answer : The final volume of gas will be, 26.3 mL
Explanation :
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is,
where,
= initial pressure of gas = 0.974 atm
= final pressure of gas = 0.993 atm
= initial volume of gas = 27.5 mL
= final volume of gas = ?
= initial temperature of gas = 
= final temperature of gas = 
Now put all the given values in the above equation, we get:
Therefore, the final volume of gas will be, 26.3 mL
ΔG° at 450. K is -198.86kJ/mol
The following is the relationship between ΔG°, ΔH, and ΔS°:
ΔH-T ΔS = ΔG
where ΔG represents the common Gibbs free energy.
the enthalpy change, ΔH
The temperature in kelvin is T.
Entropy change is ΔS.
ΔG° = -206 kJ/mol
ΔH° equals -220 kJ/mol
T = 298 K
Using the formula, we obtain:
-220kJ/mol -T ΔS° = -206kJ/mol
220 kJ/mol +206 kJ/mol =T ΔS°.
-T ΔS = 14 kJ/mol
for ΔS-14/298
ΔS=0.047 kJ/mol.K
450K for the temperature Completing a formula with values
ΔG° = (450K)(-0.047kJ/mol)-220kJ/mol
ΔG° = -220 kJ/mol + 21.14 kJ/mol.
ΔG°=198.86 kJ/mol
Learn more about ΔG° here:
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With all of the information given (pressure, volume, temperature, and the molar mass), we need a formula that relates this all together. The formula we need is the ideal gas law, PV=nRT. Since the pressure is defined in millimeters of mercury, we need the R value that correlates with this, which is 62.4; on top of this, we need the temperature in Kelvin - simply add 273.15 to convert from Celsius. With all of this information, simply plug-and-chug:
PV=nRT
(800)(3.7) = n(62.4)(37 + 273.15)
n = 0.1529 moles
Finally, the problem is asking the amount of air in grams. We have moles, so all we need to do is multiply that value by the molar mass.
0.1529 moles x 29 grams per mole =
4.435 grams of air
The balloon has 4.435 grams of air inside it.
Hope this helps!
1000 is the answer to the queshtion
