What happens to the density of gas when it compressed

Physics

1 answer:

Stella [2.4K]3 years ago

3 0

When a gas is compressed, its particles are pushed closer together so it is more dense.

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A rubber balloon is filled with 1 L of air at 1 atm and 300 K and is then put into a cryogenic refrigerator at 100 K. The rubber

Kipish [7]

Answer:

The correct answers are

(a) It decreases to 1/3 L

(ii) is (c) It is constant

Explanation:

to solve this, we list out the number of knowns and unknowns so as to determine the correct equation to solve the problem

The given variables are as follows

Initial volume V1 = 1L

V2 = Unknown

Initial Temperature T1 = 300K

let us assume that the balloon is perfectly elastic

At 300K the balloon is filled and it stretches to maintain 1 atmosphere

at 100K the content of the balloon cools reducing the excitement of the gas content which also reduces the pressure, however, the balloon being perfectly elastic, contracts to maintain the 1 atmospheric pressure, hence the answer to (ii) is (c) It is constant,

For (i) since we know that the pressure of the balloon is constant

by Charles Law V1/T1 =V2/T2

or V2 = (V1/T1)×T2 = $\frac{1L}{300K}$ × $100K$ = $\frac{1}{3}$ × L = L/3 hence the correct answer to (i) is 1/3L

8 0

3 years ago

Light strikes a 5.0-cm thick sheet of glass at an angle of incidence in air of 50°. the sheet has parallel faces and the glass h

mestny [16]

<span>Answer: sin(incidence)/sin(refraction) = n_refraction/n_incidence sin(50) / sin(x) = 1.5 / 1 sin(50)/1.5 = sin(x) sin(x) = 0.511 x = 30.71o B] 50 degrees, same as the angle going in. You can show that by reversing the steps in A. sin(30.7)/sin(x) = 1/1.5 C] The glass is 5 cm thick. The reference angle = 30.7o Tan(30.7) = displacement / thickness Tan(30.7) = x / 5 5*sin(30.7) = x x = 2.97 cm which is the displacement.</span>

4 0

3 years ago

Can you find the magnetic field based on force? a straight segment of wire 35.0 cm long carrying a current of 1.40 a is in a uni

Lostsunrise [7]

The force exerted by a magnetic field on a wire carrying current is:
$F=ILB \sin \theta$
where I is the current, L the length of the wire, B the magnetic field intensity, and $\theta$ the angle between the wire and the direction of B.

In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is $53 ^{\circ}$ , so we can re-arrange the formula and substitute the numbers to find B:
$B= \frac{F}{IL \sin \theta}= \frac{0.20 N}{(1.40 A)(0.35 m)(\sin 53^{\circ})}=0.51 T$

3 0

2 years ago

A cubic metal box with sides of 17 cm contains air at a pressure of 1 atm and a temperature of 278 K. The box is sealed so that

const2013 [10]

Answer:

F = 3.98 kN

Explanation:

GIVEN DATA:

sides of box = 17 cm

pressure = 1 atm = 101325 N/m2

T2 = 378K

T1 = 278 K