What is the average velocity of atoms in 1.00 mol of neon (a monatomic gas) at 288k?

1 answer:

7 0

You could solve this in two ways: using the ideal gas law (Van der Waals parameters optional), or by using the density. Since you specify a pressure and temperature (in kelvin), I will use the ideal gas law. Ideal gas law: PV=nRT P = pressure = 1.00 ATM V = volume = 1.00 L n = moles = what you're solving for R = gas constant = 0.0821 L*ATM/(mol*T) (T is absolute temperature (kelvin)) T = absolute temperature = 298 K (1.00atm)(1.00L) = n(0.0821L*ATM*mol -1 T -1 )(298K) n = 0.04 moles A n = Avogadro's Number = number of molecules in one mole = 6.022141 * 10 23 0.04 * 6.022141*10 23 = 2.409 * 10 22 molecules of N 2 in 1.00L at 1.00atm and 298K

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A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when

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Answer:

The speed of the resistive force is 42.426 m/s

Explanation:

Given;

mass of skydiver, m = 75 kg

terminal velocity, $V_T = 60 \ m/s$

The resistive force on the skydiver is known as drag force.

Drag force is directly proportional to square of terminal velocity.

$F_D = kV_T^2$

Where;

k is a constant

$k = \frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2}$

When the new drag force is half of the original drag force;

$F_D_2 = \frac{F_D_1}{2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_2}{V_{T2}^2} \\\\\frac{F_D_1}{V_{T1}^2} = \frac{F_D_1}{2V_{T2}^2} \\\\\frac{1}{V_{T1}^2} = \frac{1}{2V_{T2}^2}\\\\2V_{T2}^2 = V_{T1}^2\\\\V_{T2}^2= \frac{V_{T1}^2}{2} \\\\V_{T2}= \sqrt{\frac{V_{T1}^2}{2} } \\\\V_{T2}= \frac{V_{T1}}{\sqrt{2} } \\\\V_{T2}= 0.7071(V_{T1})\\\\V_{T2}= 0.7071(60 \ m/s)\\\\V_{T2}= 42.426 \ m/s$