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Damm [24]
3 years ago
9

Helium-oxygen mixtures are used by divers to avoid the bends and are used in medicine to treat some respiratory ailments. What p

ercent (by moles) of He is present in a helium-oxygen mixture having a density of 0.518 g/L at 25 ∘ C and 721 mmHg ?
Physics
1 answer:
Temka [501]3 years ago
6 0

Answer:

The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

Explanation:

From General gas equation.

PV = nRT...............................  Equation 1

Where n = number of moles, V = volume, P = pressure, T = temperature, P = pressure, V = volume.

n = mass/molar mass .................. Equation 2

substituting equation 2 into equation 1.

PV = (mass/molar mass)RT

⇒ Mass/molar mass = PV/RT..................... Equation 3

But mass = Density × Volume

⇒ M = D × V.................... Equation 4

Where D = density, M = mass

Substituting equation 4 into equation 3

DV/molar mass = PV/RT............ Equation 5

Dividing both side of the equation by Volume (V) in Equation 5

D/molar mass = P/RT .............. Equation 6

Cross multiplying equation 6

D × RT = P × molar mass

∴ Molar mass = (D × RT)/P.................. Equation 7

Where D = 0.518 g/L , R = 0.0821 atm dm³/K.mol,

T = 25°C = 25 + 273 = 298 K,

P =721 mmHg = (721/760) atm= 0.949 atm

Substituting these values into equation 7

Molar mass = (0.518 × 0.0821 × 298)/0.949

Molar mass = 13.35 g/mole

The molar mass of the mixture is =13.35 g/mole

Let y be the mole fraction of Helium and 1-y be the mole fraction of oxygen.

∴ 13.35 = 4(y) + 32(1-y)

13.35 = 4y + 32 - 32y

Collecting like terms in the equation,

32y - 4y = 32 - 13.35

28y = 18.65

y = 18.65/28

y =0.666

y = 0.666 × 100 = 66.6%

∴The percentage by mole of Helium present in the Helium-Oxygen mixture is = 66.6%

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<h3><u>Full question:</u></h3>

A generator consists of a 18-cm by 12-cm rectangular loop with 500 turns of wire spinning at 60 Hz in a 25 mT uniform magnetic field. The generator output is connected to a series RC circuit consisting of a 150 Ω and a 35 μF capacitor.What is the average power delivered to the circuit?

<h3><u>Answer:</u></h3>

The average power delivered to the circiut is 27.5 W

<h3><u>Solution:</u></h3>

Induced emf amplitude = N A B w  

N- Number of turns of the coil

B- Magnetic field

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\begin{aligned}&X_{c}=\frac{1}{(2 p i f C)}=\frac{1}{2 \times \pi \times 60 \times 35 \times 10^{-6}}\\&X_{c}=75.8\ \mathrm{ohm}\end{aligned}

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\begin{aligned}&z=\sqrt{\left[R^{2}+X_{c}^{2}\right]}=168 \text { ohm }\\&I_{\text {peak}}=\frac{101.8}{168}=0.606\ \mathrm{A}\\&P_{a v g}=\frac{l^{2} R}{2}=27.5\ \mathrm{W}\end{aligned}

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5702.88 J or 5.7mJ

Explanation:

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Where Qt = Q1+Q2

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V = 437/10

V = 43.6volts

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