Ask question

Ask question

All categories

3 years ago

10

Oxygen and hydrogen gas are at the same temperature T.What is the ratio of kinetic energies of oxygen molecule and hydrogen mole

cule if oxygen is 16 times heavier than hydrogen.

1 answer:

VashaNatasha [74]3 years ago

6 0

Answer

the ratio of kinetic energies of oxygen molecule and hydrogen molecule if oxygen is 16 times heavier than hydrogen is 1

Explanation

The average kinetic energy of a gas molecule at temperature T is given by following formula

$E =\frac{3kT}{2}$ ...............(1)

Here k is Boltzmann's constant ( $k=1.38\times 10^{-23} JK^{-1}$ ).

From above equation it is clear that average kinetic energy of a gas molecule only depends on temperature.

Therefore kinetic energy of oxygen molecule $E_{1}=\frac{3kT}{2}$ ....2

and kinetic energy of hydrogen molecule $E_{2}=\frac{3kT}{2}$ ....3

the ratio of kinetic energies of oxygen molecule and hydrogen molecule if oxygen is 16 times heavier than hydrogen is given as

$\frac{E_{1}}{E_{2}} =\frac{\frac{3kT}{2}}{\frac{3kT}{2}}$

$\frac{E_{1}}{E_{2}}=1$

Therefore, the ratio of kinetic energies of oxygen molecule and hydrogen molecule if oxygen is 16 times heavier than hydrogen is 1

You might be interested in

PLZZZ HELP 100 POINTS WILL MARK BRAINELST

Kitty [74]

Explanation:

Crust...molten

a. Oceanic, iron

b. Continental, silicates

c. less

3. Mantle, Denser

a. Lithosphere

b. Asthenosphere

4. Core

a. elements, rocks

b. liquid, magnetic

(I guess the liquid should come after the is)

Couldn't answer all but wanted to help

3 0

3 years ago

Read 2 more answers

scientists use critical thinking skills throughout the process of research or information a student noticed that their energy le

Ganezh [65]

The critical thinking step has the student just completed in the situation describe when the student concluded that soda intake really does have an effect on their energy level is the analysis. This is because the student observed the effects of drinking soda by his experience.

5 0

3 years ago

What is helpful in calculating standard deviation?

Oksana_A [137]

<span>Standard deviation is a calculation. It I used in statistical analysis of a easier job. hoped this helps u </span>

4 0

3 years ago

He shoots the tree stump, which has a mass of (M), with a bullet of mass (m) traveling at some velocity (vbullet), and the bulle

Ymorist [56]

Answer:

Explanation:

Given

mass of tree stump is $M$

mass bullet is $m$

velocity of bullet is $v$

Conserving momentum for bullet and tree stump

Initial Momentum $P_i=mv$

Suppose $v_0$ is the velocity of the system

Final Momentum $P_f=(M+m)v_0$

Initial momentum =Final Momentum

$mv=(m+M)v_0$

$v_0=\frac{mv}{m+M}$

4 0

3 years ago

A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa

Nataly_w [17]

Answer:

$V_p = 267.258 m/s$

$V_n = 38.375 m/s$

Explanation:

using the law of the conservation of the linear momentum:

$P_i = P_f$

where $P_i$ is the inicial momemtum and $P_f$ is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

$P_i = m(270 m/s)$

$P_f = mV_P + M_nV_n$

where m is the mass of the proton and $V_p$ is the velocity of the proton after the collision, $M_n$ is the mass of the nucleus and $V_n$ is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m $V_p$ + 14m $V_n$

then, m is cancelated and we have:

270 = $V_p$ + $14V_n$

This is a elastic collision, so the kinetic energy K is conservated. Then:

$K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2$

and

Kf = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

then,

$\frac{1}{2}m(270)^2$ = $\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2$

here we can cancel the m and get:

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

now, we have two equations and two incognites:

270 = $V_p$ + $14V_n$ (eq. 1)

$\frac{1}{2}(270)^2$ = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$

in the second equation, we have:

36450 = $\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2$ (eq. 2)

from this last equation we solve for $V_n$ as:

$V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

and replace in the other equation as:

270 = $V_p +$ 14 $\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }$

so,

$V_p = -267.258 m/s$

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

$V_n = \frac{270+267.258}{14}$

$V_n = 38.375 m/s$

3 0

3 years ago