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2 years ago

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A gaseous air‑fuel mixture in a sealed car engine cylinder has an initial volume of 600.mL at 1.0atm. To prepare for ignition of

the fuel, a piston moves within the cylinder, reducing the volume of the air‑fuel mixture to 50.mL at constant temperature. Assuming ideal behavior, what is the new pressure of the air‑fuel mixture?

1 answer:

Ket [755]2 years ago

7 0

12 atm is the new pressure of the air‑fuel mixture when A gaseous air‑fuel mixture in a sealed car engine cylinder has an initial volume of 600 ml at 1.0atm and final volume of 50 ml.

Explanation:

Data given:

The air fuel mixture is assumed to be having ideal behaviour

initial volume of gaseous air fuel mixture V1 = 600 ml

initial pressure of gaseous air fuel mixture P1= 1 atm

final volume when piston is removed, V2= 50 ml

final pressure of the gaseous air mixture, P2 = ?

Applying the Boyle's Law,

P1V1 = P2V2

rearranging the equation:

P2 = $\frac{P1V1}{V2}$

putting the value in the equation,

P2 = $\frac{600 X 1}{50}$

P2 = 12 atm

the pressure is increased to 12 atm when volume is reduced to 50 ml.

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What is the effect of the following on the volume of 1 mol of an ideal gas? The temperature changes from 305 K to 32°C and the p

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Answer:

The volume increases by 100%.

Explanation:

<u>Step 1:</u> Data given

Number of moles ideal gas = 1 mol

Initial temperature = 305 K

Final temperature = 32°C + 273.15 = 305.15 K

Initial pressure = 2 atm

final pressure = 101 kPa = 0.996792 atm

R = gasconstant = doesn't change

V1 = initial volume

V2= the final volume

<u>Step 2: </u>Calculate volume of original gas

P*V = n*R*T

(P*V)/ T = constante

(P1 * V1) / T1 = (P2 * V2)/ T2

In this situation we have:

(2atm * V1)/ 305 = (0.996792 *V2) / 305.15

0.006557*V1 = 0.003266*V2

V2 = 2*V1

We see that the final volume is twice the initial volume. So the volume gets doubled. The volume increases by 100%.

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Assume that each atom is a sphere, and that the surface of each atom is in contact with its nearest neighbor. Determine the perc

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Answer:

The percentage of unit cell volume that is occupied by atoms in a face- centered cubic lattice is 74.05%
The percentage of unit cell volume that is occupied by atoms in a body-centered cubic lattice is 68.03%
The percentage of unit cell volume that is occupied by atoms in a diamond lattice is 34.01%

Explanation:

The percentage of unit cell volume = Volume of atoms/Volume of unit cell

Volume of sphere = $\frac{4 }{3} \pi r^2$

a) Percentage of unit cell volume occupied by atoms in face- centered cubic lattice:

let the side of each cube = a

Volume of unit cell = Volume of cube = a³

Radius of atoms = $\frac{a\sqrt{2} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{2}}{4})^3$ = $\frac{\pi *a^3\sqrt{2}}{24}$

Number of atoms/unit cell = 4

Total volume of the atoms = $4 X \frac{\pi *a^3\sqrt{2}}{24} = \frac{\pi *a^3\sqrt{2}}{6}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{2}}{6}}{a^3} =\frac{\pi *a^3\sqrt{2}}{6a^3} = \frac{\pi \sqrt{2}}{6}$ = 0.7405

= 0.7405 X 100% = 74.05%

b) Percentage of unit cell volume occupied by atoms in a body-centered cubic lattice

Radius of atoms = $\frac{a\sqrt{3} }{4}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{4})^3$ = $\frac{\pi *a^3\sqrt{3}}{16}$

Number of atoms/unit cell = 2

Total volume of the atoms = $2X \frac{\pi *a^3\sqrt{3}}{16} = \frac{\pi *a^3\sqrt{3}}{8}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{8}}{a^3} =\frac{\pi *a^3\sqrt{3}}{8a^3} = \frac{\pi \sqrt{3}}{8}$ = 0.6803

= 0.6803 X 100% = 68.03%

c) Percentage of unit cell volume occupied by atoms in a diamond lattice

Radius of atoms = $\frac{a\sqrt{3} }{8}$

Volume of each atom = $\frac{4 }{3} \pi (\frac{a\sqrt{3}}{8})^3$ = $\frac{\pi *a^3\sqrt{3}}{128}$

Number of atoms/unit cell = 8

Total volume of the atoms = $8X \frac{\pi *a^3\sqrt{3}}{128} = \frac{\pi *a^3\sqrt{3}}{16}$

The percentage of unit cell volume = $\frac{\frac{\pi *a^3\sqrt{3}}{16}}{a^3} =\frac{\pi *a^3\sqrt{3}}{16a^3} = \frac{\pi \sqrt{3}}{16}$ = 0.3401

= 0.3401 X 100% = 34.01%

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