Answer:
<em>a. 4.21 moles</em>
<em>b. 478.6 m/s</em>
<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>
Explanation:
Volume of container = 100.0 L
Temperature = 293 K
pressure = 1 atm = 1.01325 bar
number of moles n = ?
using the gas equation PV = nRT
n = PV/RT
R = 0.08206 L-atm-

Therefore,
n = (1.01325 x 100)/(0.08206 x 293)
n = 101.325/24.04 = <em>4.21 moles</em>
The equation for root mean square velocity is
Vrms = 
R = 8.314 J/mol-K
where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol
Vrms =
= <em>478.6 m/s</em>
<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>
= 
where
Voxy = root mean square velocity of oxygen = 478.6 m/s
Vnit = root mean square velocity of nitrogen = ?
Moxy = Molar mass of oxygen = 31.9 g/mol
Mnit = Molar mass of nitrogen = 14.00 g/mol
= 
= 0.66
Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>
<em>the root mean square velocity of the oxygen gas is </em>
<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>
There's a problem with the question as given. Even with a maximum projection angle of <em>θ</em> = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height <em>y</em> at time <em>t</em> would be
<em>y</em> = (40 m/s) <em>t</em> - 1/2 <em>g t</em> ²
Set <em>y</em> = 160 m, and you'll find that there is no (real) solution for<em> t</em>, so the ball never attains the given maximum height.
From another perspective: recall that
<em>v </em>² - <em>v</em>₀² = 2<em>a </em>∆<em>y</em>
where
• <em>v</em>₀ = initial velocity
• <em>v</em> = final velocity
• <em>a</em> = acceleration
• ∆<em>y</em> = displacement
At its maximum height, the ball has zero vertical velocity, and ∆<em>y</em> = maximum height = 160 m. The ball is in free fall once it's launched, so <em>a</em> = -<em>g</em>.
So we have
0² - (40 m/s)² = -2<em>g </em>(160 m)
but this reduces to
(40 m/s)² = 2 (9.8 m/s²) (160 m)
1600 m²/s² ≠ 3136 m²/s²
As the amplitude of a sound wave increases the pitch of the ringing would be much higher (like if you were to inhale helium.. just with a phone)