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Katarina [22]
3 years ago
6

A rigid cylinder with a movable piston contains a sample of helium gas. The temperature of the gas is held

Chemistry
2 answers:
Varvara68 [4.7K]3 years ago
8 0
<span>Assuming ideal gas, we can label this system to follow the Boyle's Law. This Law can be applied to systems held at constant temperature. The formula used is: PV=k, where k is a constant. From the formula, P and V are inversely proprotional. So, if you graph P vs V, the graph would start from the top, then curves down towards the right. </span>
S_A_V [24]3 years ago
6 0

The graph that shows the relationship between the volume and pressure of the gas is \boxed{{\text{option (1)}}}. For the graph, refer to the attached image.

Further Explanation:

An ideal gas is a hypothetical gas that is composed of a large number of randomly moving particles that are supposed to have perfectly elastic collisions among themselves. It is just a theoretical concept and practically no such gas exists. But gases tend to behave almost ideally at a higher temperature and lower pressure. Helium is an example of an ideal gas.

Given information:

The temperature of the gas is kept constant.

To identify:

Graph of the relationship between pressure and volume of gas.

Boyle’s law:

It states that the volume of the gas is inversely proportional to its pressure if the temperature and the number of moles of gas remain constant. Higher the pressure of the gas, lower will be volume occupied by the gas and vice-versa. The mathematical form of Boyle’s law is,

\boxed{{\text{P}} \propto \frac{1}{{\text{V}}}}

Or,

{\text{PV}}={\text{k}}

Here,

V is volume occupied by the gas.

P is the pressure of the gas.

k is a constant.

The graph between the pressure and volume of the helium gas is drawn in the attached image. So option (1) is the correct answer.

Learn more:

1. Which statement is true for Boyle’s law: brainly.com/question/1158880

2. Calculation of volume of gas: brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: Boyle’s law, P, V, k, pressure of gas, volume occupied by gas, constant, temperature, ideal gas.

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a. i. 8.447 × 10⁻³ T ii.  27.14 cm

b. i. 2.14 cm ii. It is easily detectable.

Explanation:

a.

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Since the magnetic force F = Bqv equals the centripetal force F' = mv²/r on the C12 charge, we have

F = F'

Bqv = mv²/r

B = mv/re where B = strength of magnetic field, m = mass of C12 isotope = 1.99 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q = charge on C 12 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron)and r = radius of semicircle = 25.0 cm/2 = 12.5 cm = 12.5 × 10⁻² m

So,

B = mv/rq

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B = 16.915 × 10⁻²³ kgm/s ÷ (20.025 × 10⁻²¹ mC)

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(ii) What is the diameter of the 13C semicircle?

Since the magnetic force F = Bq'v equals the centripetal force F' = mv²/r' on the C13 charge, we have

F = F'

Bq'v = mv²/r'

r' = mv/Be where r = radius of semicircle, B = strength of magnetic field = 8.447 × 10⁻³ T, m = mass of C12 isotope = 2.16 × 10⁻²⁶ kg, v = speed of C 12 isotope = 8.50 km/s = 8.50 × 10³ m/s, q' = charge on C 13 isotope = e = electron charge = 1.602 × 10⁻¹⁹ C (since the isotope loses one electron) and  = d/2 = 12.5 cm = 12.5 × 10⁻² m

So, r' = mv/Be

r' = 2.16 × 10⁻²⁶ kg × 8.50 × 10³ m/s ÷ (8.447 × 10⁻³ T × 1.602 × 10⁻¹⁹ C)

r' = 18.36 × 10⁻²³ kgm/s ÷ 13.5321 × 10⁻²² TC)

r' = 1.357 × 10⁻¹ kgm/TC)

r' = 0.1357 m

r' = 13.57 cm

Since diameter d' = 2r', d' = 2(13.57 cm) = 27.14 cm

b.

i. What is the separation of the C12 and C13 ions at the detector at the end of the semicircle?

Since the diameter of the C12 isotope is 25.0 cm and that of the C 13 isotope is 27.14 cm, their separation at the end of the semicircle is 27.14 cm - 25.0 cm = 2.14 cm

ii. Is this distance large enough to be easily observed?

This distance of 2.14 cm easily detectable since it is in the centimeter range.

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