In a sound wave, the maximum displacement of the air molecule from it's undisturbed position is called ______________

Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy

Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

$P_{1}V_{1}=nRT$

$V_{1}=\dfrac{nRT}{P_{1}}$

Put the value into the formula

$V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}$

$V_{1}=2.62\times10^{-3}\ m^3$

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

$T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}$

Put the value into the formula

$T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}$

$T_{2}=233.7\ K$

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

$P_{1}V_{1}=P_{2}V_{2}$

$V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}$

Put the value into the formula

$V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}$

$V_{2}=0.00786\ m^3$

$V_{2}=7.86\times10^{-3}\ m^3$

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

$V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}$

$V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}$

$V_{2}=0.0057\ m^3$

$V_{2}=5.7\times10^{-3}\ m^3$

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is $7.86\times10^{-3}\ m^3$

(d). The final volume of the gas in the cylinder B is $5.7\times10^{-3}\ m^3$

6 0

3 years ago

A 2,500 kg car moves at 15 m/s^2; 15m/s^2 is the (blank) of the car.

Neporo4naja [7]

A. Acceleration.
acceleration is m/s^2. speed is m/s

7 0

3 years ago

A fluid flows along the x axis with a velocity given by V = ( x / t ) ˆ i , where x is in feet and t in seconds. (a) Plot the sp

umka21 [38]

Answer:

c)

V_local = -x/t^2

V_convec = x/t^2

d)

a = V_local + V_convec = 0

e) When a particle moves towards postive x direction its convective velocity increases, but at the same time the local velocity deacreases (at the same rate) when time increases

Explanation:

Hi!

You can see plots for a) and b) attached on this document

c)

The local acceleration is just teh aprtial derivative of the velocity with respect to t:

$\frac{dV}{dt} = \frac{d}{dt} \frac{x}{t}=- \frac{x}{t^2}$

And the convective acceleration is given by the product of the velocity times the gradient of the velocity, that is:

$\vec{v} \cdot \nabla \vec{v} = v ( \frac{dv}{dx} ) =\frac{x}{t} \frac{1}{t} = \frac{x}{t^2}$

d)

Since the acceleration of any fluid particle is the sum of the local and convective accelerations, we can easily see that it is equal to zero, since they are equal but with opposit sign

e)

This is because of teh particular form of the velocity. A particle will move towards areas of higher velocities (convectice acceleration), but as time increases, the velocity is also decreasing (local acceleration), and the sum of these quantities adds up to zero