2. Why are the health-related fitness components more

How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe

Rus_ich [418]

Explanation:

(a) For an isothermal process, work done is represented as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

Putting the given values into the above formula as follows.

W = $-nRT ln(\frac{V_{2}}{V_{1}})$

= $- 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})$

= $-12280.82 \times ln (0.09)$

= $-12280.82 \times -2.41$

= 29596.78 J

or, = 29.596 kJ (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

W = $\frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}$

= $\frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}$

= $\frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}$

= 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c) We know that for an isothermal process,

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

or, $P_{2} = \frac{P_{1}V_{1}}{V_{2}}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})$

= 11 atm

Hence, the required pressure is 11 atm.

(d) For adiabatic process,

$P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}$

or, $P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}$

= $1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}$

= 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0

3 years ago

What is the force of gravity acting on a 1-kg m mass? (g = 9.8 m/s ^ 2)

Ksenya-84 [330]

Answer: Use this F=Ma.

Explanation: So your answer will be

F=1 Kg+9.8 ms-2

So the answer will be

F=9.8N

How'd I do this?

I just used Newton's second law of motion.

I'll also put the derivation just in case.

Applied force α (Not its alpha, proportionality symbol) change in momentum

Δp α p final- p initial

Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)

or then

F α m(v-u)/t

So, as we know v=final velocity & u= initial velocity and v-u/t =a.

So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.

So, F=<em>k</em>ma (where k is the proportionality constant)

<em>k</em> is 1 so you can ignore it.

So, our equation becomes F=ma

7 0

2 years ago

Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs

Temka [501]

The motion of the ball on the vertical axis is an accelerated motion, with acceleration
$a=g=-9.81 m/s^2$
The following relationship holds for an uniformly accelerated motion:
$2aS=v_f^2 - v_i^2$
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
$v_f =0$
So we can rewrite the equation as
$2(-9.81 m/s^2) h=-v_i^2$
from which we can isolate h
$h= \frac{v_i^2}{19.62}$ (1)

Now let's assume that $v_i$ is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: $2v_i$ . So the maximum height of the second ball is
$h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62}$ (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.

8 0

2 years ago

A sample of nitrogen gas has a volume of 5.0 ml at a pressure of 1.50 atm. what is the pressure exerted by the gas if the volume

olchik [2.2K]

A sample of nitrogen gas has a volume of 5.0 ml at a pressure of 1.50 atm. what is the pressure exerted by the gas if the volume increases to 30.0 ml, at constant temperature is 0.25atm.

On constant temperature, the pressure and volume relation become constant before and after the change in quantitities have occurred.

According to Boyle's Law,

P₁V₁ = P₂V₂

where, P₁ is pressure exerted by the gas initially

V₁ is the volume of gas initially

P₂ is pressure exerted by the gas finally

V₂ is the volume of gas finally

Given,

P₁ = 1.5 atm

V₁ = 5 ml

V₂ = 30 ml

P₂ =?

On substituting the given values in the above equation:

P₁V₁ = P₂V₂

1.5 atm × 5 ml = P₂ × 30 ml

P₂ = 0.25 atm

Hence, pressure exerted by the gas is 0.25atm.

Learn more about Boyle's Law here, brainly.com/question/1437490

#SPJ4

8 0

1 year ago

Review. As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking always st

anyanavicka [17]

To find the mass of the planet we will apply the relationship of the given circumference of the planet with the given data and thus find the radius of the planet. From the kinematic equations of motion we will find the gravitational acceleration of the planet, and under the description of this value by Newton's laws the mass of the planet, that is,

The circumference of the planet is,

$\phi = 25.1m$