Ask question

Ask question

All categories

2 years ago

15

If 10.0 liters of oxygen at stp are heated to 512 °c, what will be the new volume of gas if the pressure is also increased to 15

20.0 mm of mercury?

1 answer:

Stels [109]2 years ago

3 0

Answer:

14.4 L

Explanation:

Initially, the gas is at stp (standard conditions), which means

$T_1 = 273 K\\p_1= 1.01 \cdot 10^5 Pa$

and its initial volume is

$V_1 = 10 L = 0.010 m^2$

Later, the gas is heated to a final temperature of

$T_2=512 C + 273 =785 K$

and the pressure is increased to

$p_2 = 1520.0 mmHg \cdot \frac{1.01\cdot 10^5 Pa}{760 mmHg}=2.02\cdot 10^5 Pa$

So we can use the ideal gas equation to find the new volume, V2:

$\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}\\V_2 = \frac{p_1 V_1 T_2}{p_2 T_1}=\frac{(1.01\cdot 10^5 Pa)(0.010 L)(785 K)}{(2.02\cdot 10^5 Pa)(273 K)}=0.0144 m^2 = 14.4 L$

You might be interested in

A diver in the pike position (legs straight, hands on ankles) usually makes only one or one-and-a-half rotations. To make two or

Korolek [52]

Answer:

from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

Explanation:

This can be explained on the basis of conservation of angular momentum.

This means the initial and the final angular velocity is conserved. Consider initial position (1)in the pike and final position in the be truck position. So there inertia's will also be different.

⇒ $I_1\omega_1 = I_2\omega_2$

$\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}$

also,

$I_1= mr_1^2$

$I_2= mr_2^2$

since, $r_2^2$

$I_2^2$

therefore,

$\omega_1^2$

So, from the above analysis we can say that the angular velocity in the later case is more than that of the former case. This means that the number of rotation made in the truck case is more than that made in pike position.

6 0

2 years ago

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T

atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.$

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

$\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.$

<em>where</em>,

k is the Coulomb's constant, having value = $\dfrac{1}{4\pi \epsilon_o}$

$\epsilon_o$ is the electrical permittivity of free space.

Also, r>>s, therefore, $\rm r^2+s^2\approx r^2.$

$\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.$

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

$\rm F=qE$

Therefore, the electric force on the charge Q due to the dipole is given by

$\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.$

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = $\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole is given by

$\rm \tau = Fs\ \sin\theta$

Since the charge Q is placed in the plane that bisects the dipole, therefore, $\theta = 90^\circ$ .

$\rm \tau = \dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}\cdot s\cdot 1=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs^2}{r^3}.$

4 0

2 years ago

The image shows the displacement of a motorboat. The data table shows the magnitudes of the components of each displacement vect

Diano4ka-milaya [45]

Rx= 3.5 km

Ry= 2.9 km

4 0

2 years ago

Read 2 more answers

Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be a

marysya [2.9K]

Answer:

The movable piston

Explanation:

Work is said to be done when a distance is been covered by a force . In this case kinetic energy will be change by an equal amount into work done.

Pushing the piston with a known mass of (m) and an accelarating rate from rest of ( a) to cover a known distance of (d).The idea of work done is been achieved and can be mathematically represented by:

Work done = Force x distance (d)
Force = mass (m) x acceleration (a)

8 0

2 years ago

How many calories are equal to one BTU? (One calorie = 4.186 J, one BTU = 1 054 J.)

I am Lyosha [343]

<h2>Option C is the correct answer.</h2>

Explanation:

We need to find how many calories is 1 BTU.

Given

1 BTU = 1054 J

1 calorie = 4.186 J

So we have

1 BTU = 4.186 x 251.79 J

1 BTU =251.79 calorie

1 BTU = 252 calorie.

Option C is the correct answer.

3 0

2 years ago