Which of the following is the name of the process scientists use to gain

An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v

maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

$T = \frac{PV}{nR}$

$T_{i} = T_{f}$

$\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}$

Where:

$P_{i}$ : is the initial pressure = 5.79 atm

$P_{f}$ : is the final pressure =?

$V_{i}$ : is the initial volume = 420 cm³

$V_{f}$ : is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

$P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm$

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

$W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J$

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!

3 0

3 years ago

Historia de baloncesto internacional resumen

xz_007 [3.2K]

Answer:

El baloncesto tuvo sus inicios en el año de 1891, en los Estados Unidos de Norte América justamente en la ciudad de Springfield, el mismo surgió como alternativa al Hockey sobre hielo, el cual solo se realizaba durante la época del invierno. La YMCA fue la primera asociación en practicar baloncesto cuando su pionero y fundador era entrenador en dicha asociación.

Además de los jugadores, el equipo esta conformado por masajistas, medicos, Entrenador, director y asistente técnico, los cuales tienen funciones específicas para prestar asistencia y dirección a los jugadores.

Explanation:

5 0

2 years ago

A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv

sergeinik [125]

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

$K_{rel}=(\gamma-1)mc^2$

$\dfrac{K_{rel}}{mc^2}=\gamma-1$

Put the value into the formula

$\gamma=\dfrac{105.7}{0.511}+1$

$\gamma=208$

We need to calculate the electron’s velocity

Using formula of velocity

$\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}$

$\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}$

$\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1$

$v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2$

Put the value into the formula

$v^2=\dfrac{1-(208)^2}{-208^2}\times c^2$

$v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}$

$v=0.9999 c\ m/s$

Hence, The electron’s velocity is 0.9999 c m/s.

6 0

2 years ago

An adult with a BMI of 25 to 30 is considered

soldier1979 [14.2K]

Option A overweight

HOPE IT HELPS!!

4 0

2 years ago

Read 2 more answers

A wheel in the shape of a flat, heavy, uniform, solid disk is initially at rest at the top of an inclined plane of height 2.00 m

olasank [31]

Answer:

Explanation:

If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.

mgh = ½mv²

v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s

However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½(½mR²)(v/R)²

2gh = v² + ½v²

2gh = 3v²/2

v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s

7 0

2 years ago