A _____ is a classification of an element based on its current status, position, or use in a document.

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

Answer: The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

Explanation:

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0

3 years ago

1) A toy car with a mass of 2.35 kg is pushed with a force of 21.6 N. What is the acceleration of the car?

Nataly_w [17]

Answer:

The acceleration of the car is 9,19 m/s2

Explanation:

We use the formula: F=m x a---> a=F/m

a=21,6N/ 2,35kg 1N is 1kgxm/s2

a=21,6 kg x m/s2 x 2,35 kg

a=9,191489362 m/s2

6 0

3 years ago

3. A diamond contains 0.090 moles of carbon What is the mass of the diamond?

Nady [450]

Answer:

1.0809

Explanation:

3 0

3 years ago

How do you identify a mineral there are six things

LenKa [72]

A good example is the mineral plagioclase. Plagioclase is a member of the feldspar group, but there is more than one type of plagioclase.

5 0

3 years ago

250 mL of a solution of calcium oxalate is the evaporated until only a residue of solid calcium

Lana71 [14]

Answer:

2.3 * 10^-5

Explanation:

Recall that the solubility of a solute is the amount of solute that dissolves in 1 dm^3 or 1000cm^3 of solution.

Hence;

Amount of calcium oxalate = 154 * 10^-3/128.097 g/mol = 1.2 * 10^-3 mols

From the question;

1.2 * 10^-3 mols dissolves in 250 mL

x moles dissolves in 1000mL

x = 1.2 * 10^-3 mols * 1000/250

x= 4.8 * 10^-3 moldm^-3

CaC2O4(s) ------->Ca^2+(aq) + C2O4^2-(aq)

Hence Ksp = [Ca^2+] [C2O4^2-]

Where;

[Ca^2+] = [C2O4^2-] = 4.8 * 10^-3 moldm^-3

Ksp = (4.8 * 10^-3)^2

Ksp = 2.3 * 10^-5

4 0

2 years ago

A _____ is a classification of an element based on its current status, position, or use in a document.​

A _____ is a classification of an element based on its current status, position, or use in a document.