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3 years ago

A vessel contains 1 mol of O2 and 2mol of He.what is the value of'Cp/Cv ' of the mixture?

1 answer:

6 0

Oxygen is diatomic, so its degree of freedom, (f1)= 5,
also its number of moles, n1= 1

Helium is monoatomic, so its degree of freedom (f2)= 3
and its number of moles given is, n2=2

Now using formula of effective degree of freedom of mixture, (f), we have:

f= (f1n1+f2n2)/(n1+n2)
= (5*1 + 3*2)/ (1+3)
=11/3
Also, from first law of thermodynamics;
U= n Cv. T = nRT(f2)
or, Cv = R. (f/2) (n & T cancel)

We know f=11/6,
substituting the value in above relation, we have:

Cv= R. 11/3*2
= R. 11/6

Also, Cp-Cv = R
or, Cp- R.(11/6)= R
or, Cp= R(11/6 )+1
= 17/6 R

Therefore, Cp/Cv = 17/11

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An object starts from rest, and accelerates at 2m/s2 for 10s. How far has it gone in that time

blagie [28]

Answer:

100m

Explanation:

$s = ut + \frac{1}{2} a {t}^{2}$

u=0;t=10sec;a=2m/s²

$s = 0(10) + \frac{1}{2}(2 \times {10}^{2} )$

s=10²;100m

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3 years ago

Calculate the force of a particle with a net charge of 170 coulombs traveling at a speed of 135 meters/second perpendicular to t

amm1812

Answer:

F=1.14N j

Explanation:

The magnitude of the magnetic force over a charge in a constant magnetic field is given by the formula:

$|\vec{F}|=|q\vec{v} \ X\ \vec{B}|=qvsin\theta$ (|)

In this case v and B vectors are perpendicular between them. Furthermore the direction of the magnetic force is:

-i X k = +j

Finally, by replacing in (1) we obtain:

$\vec{F}=(170C)(135\frac{m}{s})(5.0*10^{-5}T)=1.14N\ \hat{j}$

hope this helps!

6 0

3 years ago

Read 2 more answers

The Sun is about 150 million km from earth. how long does it take light from the sun to reach earth? (Speed of light is 3x10^8m/

madam [21]

Time =      (distance) / (speed)

Time = (150 x 10⁹ m) / (3 x 10⁸ m/s) =

   50 x 10¹ sec =

       <em>500 sec</em> = 8 min 20 sec

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3 years ago

Because so little can be found of the first rocks to form on

fgiga [73]

Answer:

B is the answer

Explanation:

0-/-<

7 0

3 years ago

A very long insulating cylinder has radius R and carries positive charge distributed throughout its volume. The charge distribut

blsea [12.9K]

Answer:

1. $E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. $E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3.The results from part 1 and 2 agree when r = R.

Explanation:

The volume charge density is given as

$\rho (r) = \alpha (1-\frac{r}{R})$

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.

1. Since the cylinder is very long, Gauss’ Law can be applied.

$\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

$\int\, da = 2\pi r h$

where ‘h’ is the length of the imaginary Gaussian surface.

$Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})$

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

$Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}$

3. At the boundary where r = R:

$E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}$

As can be seen from above, two E-field values are equal as predicted.

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3 years ago