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3 years ago

13

Six Carnot engines operating between different hot and cold reservoirs are described below. The heat energy transferred to the g

as during the isothermal expansion phase of each cycle is indicated.

1 answer:

Alla [95]3 years ago

7 0

Answer:

<u>Part -A</u>

The change in the entropy from largest to smallest is 3>2=4 >1 >5>6.

<u>Part-B</u>

Change in entropy over a complete cycle is 1 =2 = 3= 4 = 5 = 6

Explanation:

<u>Part -A</u>

The change in entropy for the reversible process that transfers the heat energy 'Q' at temperature 'T' is

$\Delta S = \frac{Q}{T}$

1)

$\Delta S_{1} = \frac{Q}{T}$

$= \frac{1000J}{600K}= 1.67J/K$

2)

$\Delta S_{2} = \frac{Q}{T}$

$= \frac{1000J}{500K}= 2J/K$

3)

$\Delta S_{3} = \frac{Q}{T}$

$= \frac{1500J}{600K}= 2.5J/K$

4)

$\Delta S_{4} = \frac{Q}{T}$

$= \frac{1000J}{500K}= 2J/K$

5)

$\Delta S_{5} = \frac{Q}{T}$

$= \frac{500J}{500K}= 1J/K$

6)

$\Delta S_{6} = \frac{Q}{T}$

$= \frac{500J}{600K}= 0.83J/K$

Therefore, The change in the entropy from largest to smallest is 3>2=4 >1 >5>6.

<u>Part-B</u>

$\Delta S = \frac{Q}{T}$

$Q_{system}=work\,\,done$

Initial and final states of complete cycle are equal. after complete the one cycle the system returns to the original state. So, the change in the entropy will be the zero

Therefore, Change in the overall cycle is 1 =2 = 3= 4 =5=6

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A syringe initially holds a sample of gas with a volume of 285 mL at 355 K and 1.88 atm. To what temperature must the gas in the

Ivahew [28]

<u>Answer:</u> The temperature to which the gas in the syringe must be heated is 720.5 K

<u>Explanation:</u>

To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law.

The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=1.88atm\\V_1=285mL\\T_1=355K\\P_2=2.50atm\\V_2=435mL\\T_2=?K$

Putting values in above equation, we get:

$\frac{1.88atm\times 285mL}{355K}=\frac{2.50atm\times 435mL}{T_2}\\\\T_2=\frac{2.50\times 435\times 355}{1.88\times 285}=720.5K$

Hence, the temperature to which the gas in the syringe must be heated is 720.5 K

8 0

3 years ago

8. Sulfur has a first ionization energy of 1000 kJ/mol. Photons of what frequency are required to ionize one mole of Sulfur?

Lynna [10]

Answer:

the frequency of photons $v = 1.509\times10^{39}Hz$

Explanation:

Given: first ionization energy of 1000 kJ/mol.

No. of moles of sulfur = 1 mole

$\Delta E_1 = 1000KJ/mol$

We know that plank's constant

$h = 6.626\times10^{-34} Js$

Let the frequency of photons be ν

Also we know that ΔE = hν

this implies ν = ΔE/h

$= \frac{10^6J}{6.626\times10^{-34} Js}$

$v = 1.509\times10^{39}Hz$

Hence, the frequency of photons $v = 1.509\times10^{39}Hz$

6 0

3 years ago

Which of the following options correctly ranks solid iodine, liquid bromine, and chlorine gas in order of increasing intermolecu

tatuchka [14]

Answer:

solid iodine chlorine gas

7 0

3 years ago

A certain half-reaction has a standard reduction potential E0red = +0.13V . An engineer proposes using this half-reaction at the

Ivan

Answer:

a. 1.23 V

b. No maximum

Explanation:

Required:

a. Is there a minimum standard reduction potential that the half-reaction used at the cathode of this cell can have?

b. Is there a maximum standard reduction potential that the half-reaction used at the cathode of this cell can have?

The standard cell potential (E°cell) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E°cell = E°red, cat - E°red, an

If E°cell must be at least 1.10 V (E°cell > 1.10 V),

E°red, cat - E°red, an > 1.10 V

E°red, cat - 0.13V > 1.10 V

E°red, cat > 1.23 V

The minimum standard reduction potential is 1.23 V while there is no maximum standard reduction potential.

4 0

3 years ago

A person tries to heat up her bath water by adding 5.0 l of water at 80c to 60 l of water at 30c. what is the final temperature

Sophie [7]

We can calculate the final temperature from this formula :

when Tf = (V1* T1) +(V2* T2) / (V1+ V2)

when V1 is the first volume of water = 5 L

and V2 is the second volume of water = 60 L

and T1 is the first temperature of water in Kelvin = 80 °C +273 = 353 K

and T2 is the second temperature of water in Kelvin = 30°C + 273= 303 K

and Tf is the final temperature of water in Kelvin

so, by substitution:

Tf = (5 L * 353 K ) + ( 60 L * 303 K) / ( 5 L + 60 L)

= 1765 + 18180 / 65 L

= 306 K
= 306 -273 = 33° C

8 0

3 years ago