Query3PhysRev

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course Phy 232

Query3Phys#$&*

course Phy 232

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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:

Considering PV=nRT, we know that when the temperature and volume remains constant, the number of moles and the pressure most remain constant.

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.

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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:

If only temp and volume change, T/V would have to stay constant because of the equation PV=nRT

T/V=P/(nR)

If nothing changes on the right side of the equation, the left has to stay constant which is T/V.

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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. **

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Question: prin phy Ch 13.26. Kelvin temperatures corresponding to 86 C, 78 F, -100 C, 5500 C and -459 F.

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Your Solution:

K=C+273

K=(F+460)5/9

I will put the answers in comma seperated form below

359,299,173,5773,5/9

All the answers are temperatures and they were switched from either celsius or farenheight to kelvin.

After doing this I saw that it said princ phys and not univ phys.

confidence rating #$&*:3

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Given Solution:

The Kelvin temperature is 273 K higher than the Celsius temperature (actually 273.15 below, but the degree of precision in the given temperatures is not sufficient to merit consideration of tenths or hundredths of a degree).

· 86 C, -100 C and 5500 C are therefore equivalent to ( 86 + 273 ) K = 359 K, -100 + 273 K = 173 K, (5500 + 273) K = 5773 K.

The freezing point of water is 0 C or 32 F, and a Fahrenheit degree is 5/9 the size of a Celsius degree. Therefore

· 78 F is (78 F - 32 F) = 46 F above the freezing point of water.

· 46 Fahrenheit degrees is the same as (5/9 C / F ) * 46 F = 26 C above freezing.

· Since freezing is at 0 C, this means that the temperature is 26 C.

· The Kelvin temperature is therefore (26 + 273) K = 299 K.

Similar reasoning can be used to convert -459 F to Celsius

· -459 F is (459 + 32) F = 491 F below freezing, or (5/9 C / F) * (-491 F) = 273 C below freezing.

· This is -273 C or (-273 + 273) K = 0 K.

· This is absolute zero, to the nearest degree.

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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:

Univ Phys Student

N/A

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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.

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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:

N/A

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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, and

n2 = 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.

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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:

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dT in Celsius is (150-50) = 100 C increase

50 C => (50 + 273.15) = 323.15 K

150 C => (150 + 273.15) = 423.15 K

dT in Kelvin is (432.15-323.15) = 100 K increase

50 C => (50*(9/5)+32) = 122 F

150 C => (150*(9/5)+32) = 302 F

The change in temperature in Fahrenheit is (302-122) = 180 F increase

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Self-Critique Rating:

You need to do this problem.

If it isn't marked for Principles of Physics or General College Physics, you need to do it.

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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:

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

(1*.45)/273.15 = (16*.05)/T2

T2 = 486 K

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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:

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The ratio of V/T being constant would mean that the ratio for nR/P would remain constant.

This all makes sense based on PV=nRT

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Self-Critique Rating:

@&
Good, but R is itself constant and doesn't need to be mentions.

So we would just need to say that n / P remains constant.
*@

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Question: query univ phy 17.114 / 17.116 (15.106 10th edition) 1.5 * 10^11 m, 1.5 kW/m^2, sun rad 6.96 * 10^8 m.

How did you calculate the total radiation of the Sun and how did you use this result to get the radiation per unit area?

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Your Solution:

First, by knowing or calculating the surface area, we can multiply this number by the amount radiated per unit area. The result is the total radiation of the sun.

sa= 4*pi*r^2=6.09*10^18m^2

Then, we multiply the sa by the kw/m^2 to get total radiation per unit area.

6.09*10^18m^2*1.5kW/m^2=9.13*10^18kW

confidence rating #$&*:3

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Given Solution:

Outline of solution strategy:

If we multiply the number of watts per unit of area by the surface area of the Sun we get the number of watts radiated from the Sun.

The energy flows outward in a spherically symmetric manner; at any distance the entire power is distributed over the radius of a sphere concentric with the Sun and of radius equal to the distance.

So if we divide that number of watts by the area of a sphere whose radius is equal to that of the Earth’s orbit, we get the number of watts per unit of area at that distance.

This strategy is followed in the student solution given below:

Good student solution:

Surface area of sphere of radius r is 4 pi r^2; if flux intensity is I then flux = 4 pi r^2 I.

When r = 1.5 * 10^11 m, I = 1500 W / m^2, so the flux is 4 pi r^2 I = 4 pi * (1.5 * 10^11 m)^2 * 1500 W / m^2 = 4.28 * 10^26 watts.

4.28055 x 10 ^ 26 W / (4*`pi * (6.96 x 10 ^ 8 m)^2) = 4.28055 x 10 ^ 26 W / 6.08735 x 10 ^ 18 m^2 = 70318775.82 J/s/m^2 = 7.03 x 10 ^ 7 J/s/m^2

If the sun is radiating as an ideal blackbody, e = 1, then T would be found as follows:

H = `dQ/`dt = 4.28055 x 10 ^ 26 W = (4*`pi * (6.96 x 10 ^ 8 m)^2) * (1) * (5.67051 x 10^-8 W/m^2*K) * T^4

So T^ 4 = 4.28055 x 10 ^ 26 W / 6.087351 x 10 ^ 18 m^2) * 1 * (5.67051 x 10^-8 W/m^2*K)

T^4 = 1.240 * 10 ^ 15 K ^4

T = 5934.10766 K on surface of sun. **

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Question: univ phy (omitted from 12th edition, but should be worked now) was 17.115 Solar radiation of intensity 600 watts / m^2 is incident on an ice sheet. The temperature above and below the ice sheet is 0 Celsius. Assuming that 70% of the radiation is absorbed at the surface of the ice, how long take to melt a layer of ice 1.2 cm thick?

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Your Solution:

.7*600 watts/m^2=420 watts/m^2 is the radiation absorbed by the ice.

The density of ice .917g/cm^3.

.917*.012=.011*1000=11kg

330,000 Joules/kg melted ice

11*33000=3663000Joules

3663000J/420J/s=8721 seconds

confidence rating #$&*:3

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Given Solution:

** Thermal energy is not radiating in significant quantities from the ice, so only the incoming radiation needs to be considered, and as stated only 70% of that energy is absorbed by the ice..

· 70% of the incoming 600 watts/m^2 is 420 watts / m^2, or 420 Joules/second for every square meter if ice.

· Melting takes place at 0 C so there is no thermal exchange with the environment. Thus each square meter absorbs 420 Joules of energy per second.

We need to consider the volume of ice corresponding to a square meter. Having found that we can determine the energy required to melt the given thickness:

· A 1.2 cm thickness of ice will have a volume of .012 m^3 for every square meter of surface area; the mass will be close to 1000 kg/m^3, so there are about 12 kg of ice for every m^2 of surface (you can obtain a more accurate result by using the a more accurate density; the density of ice (which floats in water) is actually somewhat less than that of water).

· It takes about 330,000 Joules to melt a kg of ice at 0 C, so to melt 12 kg requires around 4,000,000 J. At 420 Joules/sec this will require roughly 10,000 seconds, or around 3 hours.

All these calculations were done mentally and are therefore approximate. You should check them yourself, using appropriately precise values of the constants, etc. **

Self-critique:OK

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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:

H = (1/2 m^2)*(1)*(5.67*10^-8 W/m^2*K^4)(298.15^4 K) = 224.0 W

Hnet = (1/2 m^2)*(1)*(5.67*10^-8)*(298.15^4-253.15^4)=107.6 W

Hnet = (1/2 m^2)*(1)*(5.67*10^-8)*(298.15^4-3.15^4)=224.0 W

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Query3PhysRev

@& Good, but R is itself constant and doesn't need to be mentions. So we would just need to say that n / P remains constant. *@

@& Good, but check my notes. *@

@&
Good, but R is itself constant and doesn't need to be mentions.

So we would just need to say that n / P remains constant.
*@

@&
Good, but check my notes.
*@