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Chapter 1: 1.4. (page 5)

Does it ever make sense to say that one object is "twice as hot" as another? Does it matter whether one is referring to Celsius or Kelvin temperatures? Explain.

Short Answer

Expert verified

No, the statement according to physics is wrong. And the Celsius and Kelvin scales cannot give a good comparison of temperatures..

Step by step solution

01

differences between heat and temperature.

There are two distinct terms- "heat" and "temperature". heat is a term for transfer of energy from one point to another from higher temperature to the lower temperature; whereas temperature is the measure of degree of hotness or coldness of the object.

02

Comparison of heat and reading it in terms of temperature

Hence one can say that object has twice the temperature than the other, since the quantities has a clear description on how they are measured

03

The values of absolute zero of Celsius and Fahrenheit scales.

If one object was 400C and another 800C then in Celsius scale one is twice the other. But if we convert into Fahrenheit scales those temperatures come to 313.15K and 353.15K. So in this respect we can say that second object is not twice the temperature of the first. Hence we can explain the contradiction that these scales cannot be measured in the same segment.

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Most popular questions from this chapter

Put a few spoonfuls of water into a bottle with a tight lid. Make sure everything is at room temperature, measuring the temperature of the water with a thermometer to make sure. Now close the bottle and shake it as hard as you can for several minutes. When you're exhausted and ready to drop, shake it for several minutes more. Then measure the temperature again. Make a rough calculation of the expected temperature change, and compare.

If you poke a hole in a container full of gas, the gas will start leaking out. In this problem, you will make a rough estimate of the rate at which gas escapes through a hole. (This process is called effusion, at least when the hole is sufficiently small.)

  1. Consider a small portion (area = A) of the inside wall of a container full of gas. Show that the number of molecules colliding with this surface in a time interval 螖tis role="math" localid="1651729685802" PA螖t/(2mvx炉), where width="12" height="19" role="math">Pis the pressure, is the average molecular mass, and vxis the average xvelocity of those molecules that collide with the wall.
  2. It's not easy to calculate vx, but a good enough approximation is (vx2炉)1/2, where the bar now represents an average overall molecule in the gas. Show that (vx2炉)1/2=kT/m.
  3. If we now take away this small part of the wall of the container, the molecules that would have collided with it will instead escape through the hole. Assuming that nothing enters through the hole, show that the number Nof molecules inside the container as a function of time is governed by the differential equation
    dNdt=鈭�A2VkTmN
    Solve this equation (assuming constant temperature) to obtain a formula of the form N(t)=N(0)e鈭�t/r, where ris the 鈥渃haracteristic time鈥� for N(and P) to drop by a factor of e.
  4. Calculate the characteristic time for gas to escape from a 1-liter container punctured by a 1-mm2? hole.
  5. Your bicycle tire has a slow leak so that it goes flat within about an hour after being inflated. Roughly how big is the hole? (Use any reasonable estimate for the volume of the tire.)
  6. In Jules Verne鈥檚 Around the Moon, the space travelers dispose of a dog's corpse by quickly opening a window, tossing it out, and closing the window. Do you think they can do this quickly enough to prevent a significant amount of air from escaping? Justify your answer with some rough estimates and calculations.

Calculate the heat capacity of liquid water per molecule, in terms of K . Suppose (incorrectly) that all the thermal energy of water is stored in quadratic degrees of freedom. How many degrees of freedom would each molecule have to have?

Even at low density, real gases don鈥檛 quite obey the ideal gas law. A systematic way to account for deviations from ideal behavior is the virial

expansion,

PV鈭�nRT(1+B(T)(V/n)+C(T)(V/n)2+鈰�)

where the functions B(T), C(T), and so on are called the virial coefficients. When the density of the gas is fairly low, so that the volume per mole is large, each term in the series is much smaller than the one before. In many situations, it鈥檚 sufficient to omit the third term and concentrate on the second, whose coefficient B(T)is called the second virial coefficient (the first coefficient is 1). Here are some measured values of the second virial coefficient for nitrogen (N2):

T(K)
B(cm3/mol)
100鈥�160
200鈥�35
300鈥�4.2
4009.0
50016.9
60021.3
  1. For each temperature in the table, compute the second term in the virial equation, B(T)/(V/n), for nitrogen at atmospheric pressure. Discuss the validity of the ideal gas law under these conditions.
  2. Think about the forces between molecules, and explain why we might expect B(T)to be negative at low temperatures but positive at high temperatures.
  3. Any proposed relation between P, V, andT, like the ideal gas law or the virial equation, is called an equation of state. Another famous equation of state, which is qualitatively accurate even for dense fluids, is the van der Waals equation,
    (P+an2V2)(V鈭�nb)=nRT
    where a and b are constants that depend on the type of gas. Calculate the second and third virial coefficients (Band C) for a gas obeying the van der Waals equation, in terms of aand b. (Hint: The binomial expansion says that (1+x)p鈮�1+px+12p(p鈭�1)x2, provided that |px|鈮�1. Apply this approximation to the quantity [1鈭�(nb/V)]鈭�1.)
  4. Plot a graph of the van der Waals prediction for B(T), choosing aand bso as to approximately match the data given above for nitrogen. Discuss the accuracy of the van der Waals equation over this range of conditions. (The van der Waals equation is discussed much further in Section 5.3.)

Uranium has two common isotopes, with atomic masses of 238 and 235. one way to separate these isotopes is to combine the uranium with fluorine to make uranium hexafluoride gas, UF6, then exploit the difference in the average thermal speeds of molecules containing the different isotopes. Calculate the rms speed of each molecule at room temperature, and compare them.
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