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A piston-cylinder device contains an ideal gas mixture of 3 kmol of He gas and 7 kmol of Ar gas (both gases are monatomic) at 27

lidiya [134]

Answer:

Q = 62 ( since we are instructed not to include the units in the answer)

Explanation:

Given that:

$n_{HCl} = 3 \ kmol\\n_{Ar} = 7 \ k mol$

$T_1 = 27^0 \ C = ( 27+273)K = 300 K$

$P_1 = 200 \ kPa$

Q = ???

Now the gas expands at constant pressure until its volume doubles

i.e if $V_1 = x\\V_2 = 2V_1$

Using Charles Law; since pressure is constant

$V \alpha T$

$\frac{V_2}{V_1} =\frac{T_2}{T_1}$

$\frac{2V_1}{V_1} =\frac{T_2}{300}$

$T_2 = 300*2\\T_2 = 600$

mass of He =number of moles of He × molecular weight of He

mass of He = 3 kg × 4

mass of He = 12 kg

mass of Ar =number of moles of Ar × molecular weight of Ar

mass of He = 7 kg × 40

mass of He = 280 kg

Now; the amount of Heat Q transferred = $m_{He}Cp_{He} \delta T + m_{Ar}Cp_{Ar} \delta T$

From gas table

$Cp_{He} = 5.9 \ kJ/Kg/K\\Cp_{Ar} = 0.5203 \ kJ/Kg/K$

∴ Q = $12*5.19*10^3(600-300)+280*0.5203*10^3(600-300)$

Q = $62.389 *10^6$

Q = 62 MJ

Q = 62 ( since we are instructed not to include the units in the answer)

5 0

3 years ago

Read 2 more answers

Errors in the output voltage of an actual integrated circuit operational amplifier can be caused by : Select one:

natta225 [31]

Answer:

Option B

Explanation:

An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.

7 0

3 years ago

.a. What size vessel holds 2 kg water at 80°C such that 70% is vapor? What are the pressure and internal energy? b. A 1.6 m3 ves

vesna_86 [32]

Answer:

Part a: The volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b: The quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

Explanation:

Part a:

As per given data

m=2 kg

T=80 °C =80+273=353 K

Dryness=70% vapour =0.7

<em>From the steam tables at 80 °C</em>

Specific volume of saturated vapours=v_g=3.40527 $m^3/kg$

Specific volume of saturated liquid=v_f=0.00102 $m^3/kg$

Now the relation of total specific volume for a specific dryness value is given as

$v=v_f+x(v_g-v_f)$

Substituting the values give

$v=v_f+x(v_g-v_f)\\v=0.00102+0.7(3.40527-0.00102)\\v_f=2.38399 m^3/kg$

Now the volume of vessel is given as

$v=\frac{V}{m}\\V=v \times m\\V=2.38399 \times 2\\V=4.7680 m^3$

So the volume of vessel is 4.7680 $m^3$ .

Similarly for T=80 and dryness ratio of 0.7 from the table of steam

Pressure=P=47.4 kPa

Specific internal energy is given as u=1840 kJ/kg

So the total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=1840 \times 2\\U=3680 kJ$

The total internal energy is 3680 kJ.

So the volume of vessel is 4.7680 $m^3$ and total internal energy is 3680 kJ.

Part b

Volume of vessel is given as 1.6

mass is given as 2 kg

Pressure is given as 0.2 MPa or 200 kPa

Now the specific volume is given as

$v=\frac{V}{m}\\v=\frac{1.6}{2}\\v=0.8 m^3/kg$

So from steam tables for Pressure=200 kPa and specific volume as 0.8 gives

Temperature=T=120 °C

Quality=x=0.903 ≈ 90.3%

Specific internal energy =u=2330 kJ/kg

The total internal energy is given as

$u=\frac{U}{m}\\U=u \times m\\U=2330 \times 2\\U=4660 kJ$

So the quality of the mixture is 90.3% or 0.903, temperature is 120 °C and total internal energy is 4660 kJ.

5 0

3 years ago

What are the inputs, outputs and side effects of a PSP?

Kipish [7]

Answer:

Side effects - sudden loss of balance/ repeated falls

Outputs - sever sickness and could me factual

Inputs/corrections of this- medications and experimental treatments to help slow the process of deterioration

5 0

3 years ago

Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given

Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = $\frac{1}{6}*334.8 = 55.8 ft^3$

Volume of sand = $\frac{2}{6}*334.8 = 111.6 ft^3$

Volume of gravel = $\frac{3}{6}*334.8 = 167.4 ft^3$

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

4 0

3 years ago