Ask question

Ask question

All categories

2 years ago

3

Suppose that 10 moles of an ideal gas have a gauge pressure of 2 atm and a temperature of 200 K. If the volume of the gas is dou

bled and the pressure dropped to a gauge pressure of 1 atm, what is the new temperature?
Select one:

a.

267 K

b.

300 K

c.

400 K

d.

200 K

1 answer:

gavmur [86]2 years ago

4 0

To solve this problem we will apply the concepts related to the equations of ideal gas, from there, we will define the proportion of these values between two states of matter. The ideal gas equation is defined as

$PV = nRT$

Here,

P = Pressure

V = Volume

n = Amount of mass

R = Gas ideal constant

T = Temperature

Between two states we have,

$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$

The pressure at each state is,

$P_1 = P_{atm}+2 = 3atm$

$P_2 = P_{atm}+1 = 2atm$

According to the statement,

$V_2 = 2V_1$

Then replacing,

$T_2(3atm)(V_1) = T_1(2atm)(2V_1)$

$T_2 = \frac{4}{3}T_1$

$T_2 = \frac{4}{3} 200$

$T_2 = 267K$

Therefore the correct answer is A.

You might be interested in

A ball is launched with initial speed v from ground level up a frictionless slope (This means the ball slides up the slope witho

amid [387]

Answer:

hmax = 1/2 · v²/g

Explanation:

Hi there!

Due to the conservation of energy and since there is no dissipative force (like friction) all the kinetic energy (KE) of the ball has to be converted into gravitational potential energy (PE) when the ball comes to stop.

KE = PE

Where KE is the initial kinetic energy and PE is the final potential energy.

The kinetic energy of the ball is calculated as follows:

KE = 1/2 · m · v²

Where:

m = mass of the ball

v = velocity.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the ball.

g = acceleration due to gravity (known value: 9.81 m/s²).

h = height.

At the maximum height, the potential energy is equal to the initial kinetic energy because the energy is conserved, i.e, all the kinetic energy was converted into potential energy (there was no energy dissipation as heat because there was no friction). Then:

PE = KE

m · g · hmax = 1/2 · m · v²

Solving for hmax:

hmax = 1/2 · v² / g

4 0

2 years ago

Before leaving the house in the morning, you plop some stew in your slow cooker and turn it on Low. The slow cooker has a 160 Oh

guajiro [1.7K]

Answer:

Total charge flow through the cooker is 21600 C

Explanation:

As we know that the current flow through the cooker is given by Ohm's law

here it is given as

$V = i R$

$i = \frac{V}{R}$

$i = \frac{120}{160}$

$i = \frac{3}{4} A$

now the charge flow through it is given as

$Q = i t$

total time is t = 8 hours

$Q = \frac{3}{4}(8 \times 60 \times 60)$

$Q = 21600 C$

7 0

2 years ago

Maria throws an apple vertically upward from a height of 1.3 m with an initial velocity of +2.8 m/s. Will the apple reach a frie

evablogger [386]

Answer:

No, the apple will reach 4.20041 m below the tree house.

Explanation:

t = Time taken

u = Initial velocity = 2.8 m/s

v = Final velocity = 0

s = Displacement

g = Acceleration due to gravity = -9.81 m/s² = a (negative as it is going up)

Equation of motion

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-2.8^2}{2\times -9.81}\\\Rightarrow s=0.39959\ m$

The height to which the apple above the point of release will reach is 0.39959 m

From the ground the distance will be 1.3+0.39959 = 1.69959 m

Distance from the tree house = 5.9-1.69959 = 4.20041 m

No, the apple will reach 4.20041 m below the tree house.

The values in the option do not reflect the answer.

5 0

2 years ago

Let’s consider tunneling of an electron outside of a potential well. The formula for the transmission coefficient is T \simeq e^

ioda

Answer:

L' = 1.231L

Explanation:

The transmission coefficient, in a tunneling process in which an electron is involved, can be approximated to the following expression:

$T \approx e^{-2CL}$

L: width of the barrier

C: constant that includes particle energy and barrier height

You have that the transmission coefficient for a specific value of L is T = 0.050. Furthermore, you have that for a new value of the width of the barrier, let's say, L', the value of the transmission coefficient is T'=0.025.

To find the new value of the L' you can write down both situation for T and T', as in the following:

$0.050=e^{-2CL}\ \ \ \ (1)\\\\0.025=e^{-2CL'}\ \ \ \ (2)$

Next, by properties of logarithms, you can apply Ln to both equations (1) and (2):

$ln(0.050)=ln(e^{-2CL})=-2CL\ \ \ \ (3)\\\\ln(0.025)=ln(e^{-2CL'})=-2CL'\ \ \ \ (4)$

Next, you divide the equation (3) into (4), and finally, you solve for L':

$\frac{ln(0.050)}{ln(0.025)}=\frac{-2CL}{-2CL'}=\frac{L}{L'}\\\\0.812=\frac{L}{L'}\\\\L'=\frac{L}{0.812}=1.231L$

hence, when the trnasmission coeeficient has changes to a values of 0.025, the new width of the barrier L' is 1.231 L

8 0

2 years ago

A container of nitrogen (an ideal diatomic gas, molecular weight=28) is at a pressure of 2 atm and has a mass density of 1.6 gra

Nikolay [14]

Answer:

Explanation:What is the root-mean-square velocity (vrms) of the center of mass of one of the molecules?

3 0

2 years ago