Query 3

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course Phy 202

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it.  This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

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Question:    

query  problem 15 introductory problem sets temperature and volume information find final temperature.

 

When temperature and volume remain constant what ratio remains constant?

 

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Your Solution:

 PV = n R T so n R / P = V / T

 

 

confidence rating #$&*:

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Given Solution: 

** PV = n R T so n R / P = V / T

 

Since T and V remain constant, V / T remains constant. 

 

·          Therefore n R / P remain constant.

·          Since R is constant it follows that n / P remains constant. **

STUDENT QUESTION:

I don’t understand why P is in the denominator when nR was moved to the left side of the equation
INSTRUCTOR RESPONSE:

The given equation was obtained by dividing both sides by P and by T, then reversing the sides.
We could equally well have divided both sides by v and by n R to obtain
P / (n R) = T / V,
and would have concluded that P / n is constant.
To say that P / n is constant is equivalent to saying the n / P is constant. 

Your Self-Critique: ok

 

 

Your Self-Critique Rating: ok

 

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Question:  why, in terms of the ideal gas law, is T / V constant when only temperature and volume change?

 

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Your Solution:

Well PV=nRT so when you switch that around you get T/V=P/n/R

 

confidence rating #$&*:

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Given Solution: 

** STUDENT ANSWER AND INSTRUCTOR RESPONSE: 

 

They are inversely proportional. They must change together to maintain that proportion.

 

INSTRUCTOR RESPONSE: 

 

You haven't justified your answer in terms of the ideal gas law: 

 

PV = n R T so V / T = n R / P.

 

If only T and V change, n and P don't change so n R / P is constant.

 

Therefore V / T is constant, and so therefore is T / V.

 

You need to be able to justify any of the less general relationships (Boyle's Law, Charles' Law, etc.) in terms of the general gas law, using a similar strategy.  **

 

 

Your Self-Critique: ok

 

 

Your Self-Critique Rating: 3

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Question:    prin phy and gen phy Ch 13.30 air at 20 C is compressed to 1/9 of its original volume.  Estimate the temperature of the compressed air assuming the pressure reaches 40 atm.

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Your Solution:

 P v=nRT

T2 = (P2 / P1) * (V2 / V1) * T1.

  P2 / P1 = 40

V2 / V1 = 1/9

T2 = 40 * 1/9 * T1

original temperature is 20 C = 293 K so that T1 = 293 K 

T2 = 40 * 1/9 * 293 K

 

 

confidence rating #$&*: 2

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Given Solution: 

 

 

First we reason this out intuitively: 

 

If the air was compressed to 1/9 its original volume and the temperature didn’t change, it would end up with 9 times its original pressure. 

 

However the pressure changes from 1 atm to 40 atm, which is a 40-fold increase. 

 

The only way the pressure could end up at 40 times the original pressure, as opposed to 9 times the original, would be to heat up.  Its absolute temperature would therefore have to rise by a factor of 40 / 9.

 

Its original temperature was 20 C = 293 K, so the final temperature would be 293 K * 40/9, or over 1300 K. 

 

Now we reason in terms of the ideal gas law. 

 

P V = n R T. 

 

In this situation the number of moles n of the gas remains constant.  Thus P V / T = n R, which is constant, and thus P1 V1 / T1 = P2 V2 /T2. 

 

The final temperature T2 is therefore

 

·          T2 = (P2 / P1) * (V2 / V1) * T1.

 

From the given information P2 / P1 = 40 and V2 / V1 = 1/9 so

 

·          T2 = 40 * 1/9 * T1.

 

The original temperature is 20 C = 293 K so that T1 = 293 K, and we get

 

·          T2 = 40 * 1/9 * 293 K,

 

the same result as before.

 

 

Your Self-Critique:

 This problem was a little bit tricky for me.

 

 

Your Self-Critique Rating: 3

 

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Question:    query  gen phy ch 13.38 fraction of air released after tire temp increases from 15 to 38 C at 220 kPa gauge  

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Your Solution:

 T1 = 288 K, T2 = 311 K

T2 / T1 = 311 / 288 = 1.08 (This is about an 8% increases in temp)

P2 = 3ll / 288 * 321 kPa = 346 kPa

 ???I am unsure of what to do the rest of the problem…

 

 

confidence rating #$&*: 2

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Given Solution: 

 

(Note that the given 220 kPa initial gauge pressure implies an absolute pressure of 311 k Pa; assuming atmospheric pressure of about 101 k Pa, we add this to the gauge pressure to get absolute pressure).

Remember that the gas laws are stated in terms of absolute temperature and pressure.

The gas goes through three states.  The temperature and pressure change between the first and second states, leaving the volume and the number n of moles constant.  Between the second and third states pressure returns to its original value while volume remains constant and the number n of moles decreases. 

 

From the first state to the second:

 

T1 = 288 K, T2 = 311 K so T2 / T1 = 311 / 288 = 1.08, approx. 

 

This is approx. an 8% increase in temperature.  The pressure must therefore rise to

 

P2 = 3ll / 288 * 321 kPa = 346 kPa, approx

 

(note that we have to use actual rather than gauge pressure so init pressure is 220 kPa + 101 kPa = 321 kPa, approx.  )

 

From the second state to the third, pressure is then released by releasing some gas, changing the number n of moles of gas in order to get pressure back to 331 kPa.  Thus

 

n3 / n2 = P3 / P2 = 321 kPa / 346 kPa or approximately .93, which would be about a 7% decrease.  So we have to release about 7% of the air.

 

Note that these calculations have been done mentally, and they might not be particularly accurate.  Work out the process to botain the accurate numerical results.

 

Note also that temperature changes from the second to third state were not mentioned in the problem.  We would in fact expect a temperature change to accompany the release of the air, but this applies only to the air that escapes.  The air left in the tire would probably change temperature for one reason or another, but it wouldn't do so as a direct result of releasing the air.

STUDENT QUESTION

It seems that the air goes from 288 to 311 K, so the ratio should be n2 / n1 = 288 / 311 and the proportional loss should be about (1 - 288 / 311)

INSTRUCTOR RESPONSE

The Kelvin temperature goes from 288 K to 311 K.
If the air is released at constant pressure, then volume and pressure remain constant while temperature and number of moles vary according to 
n T = P V / R
so that
n1 T1 = n2 T2, andn2 = n1 * (T1 / T2) = n1 * (288 / 311)
and the change in amount of gas is n1 - n2 = n1 - 288/311 n1 = n1( 1 - 288 / 311), or about 7.4% of n1.
If the temperature is first raised to 311 K, then the gas is released, it is the pressure and amount of gas that change. In that case the change in the amount of the gas is n1 - n3 = n1 - (321 kPa / 346 kPa) * n1 = n1 ( 1 - 321 / 346), or about 7.2% of n1.
The fractions 288/311 and 321 / 346 don't differ by much, not do the percents, but they do differ.

 

Your Self-Critique:

 After looking over the solution, I am still slightly confused, I hope the more problems I do the more comfortable I will be with this material.

 

 

Your Self-Critique Rating: 3

@&

n3 and n2 are the numbers of moles of gas.

n3 / n2 = .93, meaning that .93 or 93% of the original gas is still in the tire.

So 7% must have been released.

*@

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Question:    `q001.  The temperature of a certain object increases from 50 Celsius to 150 Celsius.  What is its change in temperature in Celsius?

Convert 50 Celsius and 150 Celsius to Kelvin.

What is the change in temperature in Kelvin?

What is the change in temperature in Fahrenheit?

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Your Solution:

323 K (50 C)

423 K (150 C)

Change in Temp: 423 K-323 K= 100 K

Change in F= 302 F-122 F= 180 F

confidence rating #$&*:

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Self-Critique Rating: ok

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Question:    `q002.  A sample of gas originally at 0 Celsius and pressure 1 atmosphere is compressed from volume 450 milliliters to volume 50 milliliters, in which state its pressure is 16 atmospheres.  What is its temperature in the new state?

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Your Solution:

PV = nRT

Original: (1 atm)(.450 L) = n(0.08206)(273 K)

n = 0.02

New: (16 atm)(0.050 L) = (0.02)(0.08206)(new temp)

new temp = 487.4 K = 214.4 C

confidence rating #$&*:

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Self-Critique Rating: ok

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Question:    `q003.  What product or ratio involving P, V, n and T would remain constant if V and T were held constant?

Why does this make sense?

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Your Solution:

If V and T were constant, then the V/T ratio would remain constant. Also, the n/P ratio would be constant too since n and P are the only other aspects of the equation that could change.

confidence rating #$&*:

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Self-Critique Rating: ok

 

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Question: 

(a) What is the average kinetic energy in joules of hydrogen atoms on the 5500ºC surface of the Sun?

(b) What is the average kinetic energy of helium atoms in a region of the solar corona where the temperature is 6.00×10^5 K ?

The average kinetic energy of particles at temperature T is

KE_ave = 3/2 k T,

where k = R / N_A is the Boltzmann constant (R is the gas constant and N_A is avagodro's number).

The average kinetic energy of a particle does not depend on what kind of particle it is.  However the less massive the particle, the greater will be its speed at the average kinetic energy.

 

At 5500 C the absolute temperature is 5773 K so the average KE of a particle is

KE_ave = 3/2 k T = 3/2 * 1.38 * 10^-23 Joules / (particle Kelvin) * 5773 K = 1.19 * 10^-19 Joules.

 

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Your solution: 

????I have no idea how to even attempt this problem…  

 

 

confidence rating #$&*: 2

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Given Solution: 

 

 

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Self-critique (if necessary):

 

 

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Self-critique Rating: 

 

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Question:    `q004.  A body with a 1/2 m^2 surface area, at temperature 25 Celsius, has an emissivity of approximately 1.  It exchanges energy by radiation with a large surface at temperature -20 Celsius.  At what net rate does it lose energy?

By what percent would the rate of energy loss change if the large surface was at temperature -270 Celsius?

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Your Solution:

For this I can use the equation Hnet = Ae(stefan-boltzmann constant)(T^4-Ts^4)

Hnet = (1/2)(1)(5.67051 x 10^-8 W/m^2*K)((298 K)^4 - (253 K)^4) = (1/2)(1)(5.67051 x 10^-8 W/m^2*K)(3788998335 K) = 107.43 W

If the large surface was at -270 celsius, then the new rate would be:

Hnet = (1/2)(1)(5.67051 x 10^-8 W/m^2*K)((298 K)^4 - (3 K)^4) = (1/2)(1)(5.67051 x 10^-8 W/m^2*K)(7886150335 K) = 223.6 W

It would change by (223.6 W / 107.43 W) x 100 = 208.12 %

confidence rating #$&*:

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Self-Critique Rating: ok"

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