All categories

3 years ago

All of the following equations are statements of the ideal gas law except

Physics

1 answer:

elena55 [62]3 years ago

7 0

Answer:

a. P = nRTV

Explanation:

The question is incomplete. Here is the complete question.

"All of the following equations are statements of the ideal gas law except a. P = nRTV b. PV/T = nR c. P/n = RT/v d. R = PV/nT"

Ideal gas equation is an equation that describes the nature of an ideal gas. The molecule of an ideal gas moves at a particular velocity depending on the temperature. This gases collides with one another elastically. The collision that an ideal gas experience is a perfectly elastic collision.

The ideal gas equation is expressed as shown:

PV = nRT where:

P is the pressure of the gas

V is the volume

n is the number of moles

R is the ideal gas constant

T is the temperature.

Based on the formula given for an ideal gas, it can be inferred that the equation. P = nRTV is not a statement of an ideal gas equation.

The remaining option will results to an ideal gas equation if they are cross multipled.

You might be interested in

What is a continuous range of a single feature such as a wave lengt

Evgen [1.6K]

<span>I think you're looking for "spectrum".</span>

7 0

3 years ago

Newton began his academic career in 1667 for how long was he a working scientists was he a very productive scientist

taurus [48]

Newton was given the title "scientist" when he was given the Merit Badge at the age of 15. He continued to work in the field of science, and was considered a scientist until his death at the age of 84. That is a total of 69 years of work. Hope this helps!

6 0

2 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

4 0

2 years ago

NEED HELP ASAP, ILL GIVE YOU BRAINLIEST IF CORRECT (30POINTS)

Vera_Pavlovna [14]

Answer:

the bar is the top and bottem. the nucleas in the middle and the Spiral arm is the last space

Explanation:

5 0

2 years ago

In a candy factory, the nutty chocolate bars contain 20.0 % pecans by mass. If 4.0 kg of pecans were used for candy last Tuesday

ladessa [460]

Answer:

44.09 pounds

Explanation:

We got that 20 % of the mass of a nutty chocolate bar its pecans, if 4.0 kg of pecans were used, we need to find the X in the equation

$0.2 * X = 4.0 \ kg$

where X its the total mass of nutty chocolate bars produced. So, we can just divide by 0.2 on both sides, and we find:

$X = 4.0 kg / 0.2$

$X = 20.0 \ kg$

Of course, we need the total mass produced in pounds, and not in kilograms. Looking at an conversion table, we can find that 1 kg its 2.20462 pounds, multiplying the value for total mass produced by the conversion factor we get:

$X = 20.0 \ kg * 2.20462 \ \frac{pounds}{kg}$

$X = 44.0924 \ pounds$

Now, we just need to round off to two significant figures. This is:

$X = 44.09 \ pounds$ ,

the total mass of nutty chocolate bars made last Tuesday to two significant figures.

5 0

3 years ago