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10 months ago

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A monatomic ideal gas undergoes a quasi-static process that is described by the function p(????)=p1+3(????−????1) , where the st

arting state is (p1,????1) and the final state (p2,????2). Assume the system consists of n moles of the gas in a container that can exchange heat with the environment and whose volume can change freely.
(a) Evaluate the work done by the gas during the change in the state.

(b) Find the change in internal energy of the gas.

(c) Find the heat input to the gas during the change.

(d) What are initial and final temperatures?

1 answer:

Alenkasestr [34]10 months ago

5 0

A pure gas made up only of atoms. The noble gases argon, krypton, and xenon are some examples.

Concepts:

Perfect gas law: Work performed on the system: PV = nRT W = -∫PdV

Energy preservation formula: U = Q + W

Reasoning:

W = nRT ln(Vi/Vf) when the process is isothermal.

The temperature is said to be constant, and we are given n, Pfinal, and Vfinal.

Calculation information:

(A) A process that is isothermal has a constant temperature.

PV = nRT, and hence, constant

nRT = PV = 101000 Pa*25*10-3 m3

For a process that is isothermal, W = nRT ln(Vi/Vf).

W/(nRT)=3000 J/(101000 Pa*25*10-3 m3)=-1.19

(The gas produces -W of labor.)

Vi = (25*10-3 m3)/3.28 = 7.62*10-3 m3 = 7.62 L where Vf/Vi = exp(1.19) = 3.28 Vi (b) for a perfect gas PV = nRT. 101000 Pa*25*10-3 m3 = (8.31 J/K) T. T = 303.85 K.

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Answer:

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e = 0.7986

Explanation:

We know that moist unit weight of soil is given as

$\gamma_m \ or\ bulk\ density = \frac{(Gs+Se)\times \gamma_w}{(1+e)}$

where, $\gamma_m$ = moist unit weight of the soil

Gs = specific gravity of the soil

S = degree of saturation

e = void ratio

$\gamma_w$ = unit weight of water = 9.81 kN/m3

From data given we know that:

At 50% saturation, $\gamma_m = 16.62 kN/m3$

puttng all value to get Gs value;

$16.62= \frac{(Gs+0.5*e)\timees 9.81}{(1+e)}$

Gs - 1.194*e = 1.694 .........(1)

for saturaion 75%, unit weight = 17.71 KN/m3

$17.71 = \frac{(Gs+0.75*e)\times 9.81}{(1+e)}$

Gs - 1.055*e = 1.805 .........(2)

solving both equations (1) and (2), we obtained;

Gs = 2.647

e = 0.7986

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3 years ago

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Answer:

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Answer:

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Explanation:

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if:

Gsky=oTsky^4

Eb=oTs^4

qc=h(Ts-Tα)

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha l Gl} \, dl }{\int\limits^\alpha _0 {Gl} \, dl }$

$\alpha s=\frac{\int\limits^\alpha _0 {\alpha lEl(l,5800 } \, dl }{\int\limits^\alpha _0 {El(l,5800)} \, dl }$

if Gl≈El(l,5800)

$\alpha s=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

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$\alpha s=(0.8*0.941)+0.3(1-0.941)=0.77$

The hemispherical emissivity is equal to:

$E=(1-0.2)F(0-2)+(1-0.7)(1-F(0-2))$

lt=2*333=666 K, at this value, F=0

$E=0+(1-0.7)(1)=0.3$

The hemispherical absorptivity is equal to:

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3 years ago

A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with fri

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Answer:

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Therefore, tension in the cable, $T_{BE}=196.2 N$

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$24.525-39.24+0.2D_x=0$

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Similarly,

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3 years ago

Air exits a compressor operating at steady-state, steady-flow conditions at 150 oC, 825 kPa, with a velocity of 10 m/s through a

ioda

Answer:

a) Qe = 0.01963 m^3 / s , mass flow rate m^ = 0.1334 kg/s

b) Inlet cross sectional area = Ai = 0.11217 m^2 , Qi = 0.11217 m^3 / s

Explanation:

Given:-

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Pressure ( Pe ) = 825 KPa

Temperature ( Te ) = 150°C

Velocity ( Ve ) = 10 m/s

Diameter ( de ) = 5.0 cm

Solution:-

- Define inlet parameters:

Pressure = Pi = 100 KPa

Temperature = Ti = 20.0

Velocity = Vi = 1.0 m/s

Area = Ai

- From definition the volumetric flow rate at outlet ( Qe ) is determined by the following equation:

Qe = Ae*Ve

Where,

Ae: The exit cross sectional area

Ae = π*de^2 / 4

Therefore,

Qe = Ve*π*de^2 / 4

Qe = 10*π*0.05^2 / 4

Qe = 0.01963 m^3 / s

- To determine the mass flow rate ( m^ ) through the compressor we need to determine the density of air at exit using exit conditions.

- We will assume air to be an ideal gas. Thus using the ideal gas state equation we have:

Pe / ρe = R*Te

Where,

Te: The absolute temperature at exit

ρe: The density of air at exit

R: the specific gas constant for air = 0.287 KJ /kg.K

ρe = Pe / (R*Te)

ρe = 825 / (0.287*( 273 + 150 ) )

ρe = 6.79566 kg/m^3

- The mass flow rate ( m^ ) is given:

m^ = ρe*Qe

= ( 6.79566 )*( 0.01963 )

= 0.1334 kg/s

- We will use the "continuity equation " for steady state flow inside the compressor i.e mass flow rate remains constant:

m^ = ρe*Ae*Ve = ρi*Ai*Vi

- Density of air at inlet using inlet conditions. Again, using the ideal gas state equation:

Pi / ρi = R*Ti

Where,

Ti: The absolute temperature at inlet

ρi: The density of air at inlet

R: the specific gas constant for air = 0.287 KJ /kg.K

ρi = Pi / (R*Ti)

ρi = 100 / (0.287*( 273 + 20 ) )

ρi = 1.18918 kg/m^3

Using continuity expression:

Ai = m^ / ρi*Vi

Ai = 0.1334 / 1.18918*1

Ai = 0.11217 m^2

- From definition the volumetric flow rate at inlet ( Qi ) is determined by the following equation:

Qi = Ai*Vi

Where,

Ai: The inlet cross sectional area

Qi = 0.11217*1

Qi = 0.11217 m^3 / s

- The equations that will help us with required plots are:

Inlet cross section area ( Ai )

Ai = m^ / ρi*Vi

Ai = 0.1334 / 1.18918*Vi

Ai ( V ) = 0.11217 / Vi .... Eq 1

Inlet flow rate ( Qi ):

Qi = 0.11217 m^3 / s ... constant Eq 2

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3 years ago