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Korvikt [17]
3 years ago
9

Oxygen gas having a volume of 1010 cm3 at 20.7°C and 1.04 x 105 Pa expands until its volume is 1570 cm3 and its pressure is 1.07

x 105 Pa. Find (a) the number of moles of oxygen present and (b) the final temperature of the sample.
Physics
1 answer:
Ksju [112]3 years ago
4 0

Answer:

final temperature is 469.71 K = 196.71 °C

Explanation:

Given data

volume V = 1010 cm3 = 0.00101 m³

temperature = 20.7°C  = 20.7 + 273 = 293.7 K

pressure P1 = 1.04 x 10^5 Pa

volume = 1570 cm3

pressure P2 = 1.07 x 10^5 Pa

to find out

number of moles of oxygen and   final temperature

solution

we know that for ideal gas

PV = nRT

put here all value to get n

R gas constant = 8.314472 J/mol−K  and T = 293.7 K and  V = 0.00101 m³

and P = 1.04 x 10^5

so n = ( 1.04 x 10^5  × 0.00101 ) / ( 8.314472 ×  293.7 )

n = 0.043015  moles

and

now we use equation

P1V1 /T1 = P2V2 / T2

1.04 x 10^5 × 1010  / 293.7 = 1.07 x 10^5 × 1570  / T2

T2 = 49338663  / 105040 = 469.71 K

so final temperature is 469.71 K = 196.71 °C

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Answer:

The thickness of the foil is 0.017 mm.

Explanation:

Given that,

Weight = 7.5 oz = 212.6175 gm

Density = 2.70 g/cm³

Area of aluminium = 50 ft² = 46451.52 cm²

We need to calculate the thickness of the foil

Using formula of density

\rho=\dfrac{m}{V}

\rho=\dfrac{m}{A\times t}

t=\dfrac{m}{A\times \rho}

Where, A = area

t = thickness

m = mass

Put the value into the formula

t=\dfrac{212.6175}{46451.52\times2.70}

t=0.00170\ cm

t=0.017\ mm

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Answer:

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lutik1710 [3]

Answer:

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Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

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Where T is the period of revolution and a is the semi-major axis.

In the other hand, the distance between the Earth and the Sun has a value of 1.50x10^{8} Km. That value can be known as well as an astronomical unit (1AU).

But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.

For the special case of the asteroid the distance will be:

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That distance will be expressed in terms of astronomical units:

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Finally, from equation 1 the period T can be isolated:

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Answer:

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