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A full 15.0 L scuba tank at 295 K contains 144 moles of air and has a pressure of 234 atm. At the end of the dive, the tank has a pressure of 68.0 atm and a temperature of 280 K. How many moles of air are left in the tank?"

A Full 150 L Scuba Tank At 295 K Contains 144 Moles Of Air And Has A Pressure Of 234 Atm At The End Of The Dive The Tank Has A Pressure Of 680 Atm And A Tempera class=

Sagot :

CrystalQueenidn

Answer:

44.1 L

Explanation:

Since volume is being held constant, we can use the following variation of the Ideal Gas Law to find the new pressure.

[tex]\frac{P_1}{T_1N_1}=\frac{P_2}{T_2N_2}[/tex]

In the equation, "P₁", "T₁", and "N₁" represent the initial pressure, temperature, and moles. "P₂", "T₂", and "N₂" represent the final pressure, temperature, and moles. Your answer should have 3 sig figs to match the sig figs of the given values.

P₁ = 234 atm                                 P₂ = 68.0 atm

T₁ = 295 K                                     T₂ = 280 K

N₁ = 144 moles                              N₂ = ? moles

[tex]\frac{P_1}{T_1N_1}=\frac{P_2}{T_2N_2}[/tex]                                              <----- Equation

[tex]\frac{234 atm}{(295 K)(144 moles)}=\frac{68.0 atm}{(280 K)N_2}[/tex]                        <----- Insert values

[tex]\frac{234 atm}{42480}=\frac{68.0 atm}{(280 K)N_2}[/tex]                                      <----- Multiply 295 and 144

[tex]0.00551=\frac{68.0 atm}{(280 K)N_2}[/tex]                                    <----- Simplify left side

[tex]1.54=\frac{68.0 atm}{N_2}[/tex]                                            <----- Multiply both sides by 280

[tex](1.54)N_2={68.0 atm}[/tex]                                   <----- Multiply both sides by N₂

[tex]N_2 = 44.1L[/tex]                                                <----- Divide both sides by 1.54

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