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Chapter 6: Q. 6.22P. (page 234)

In most paramagnetic materials, the individual magnetic particles have more than two independent states (orientations). The number of independent states depends on the particle's angular momentum 鈥渜uantum number鈥� j, which must be a multiple of 1/2. For j = 1/2 there are just two independent states, as discussed in the text above and in Section 3.3. More generally, the allowed values of the z component of a particle's magnetic moment are

Short Answer

Expert verified

The expression of Newton's law of motion is as follows,

Here is the mass of the object and is the acceleration of the object.

Free body diagram of the Rocket is as follows,

Step by step solution

01

Step 1. Given.

Apply the Newton's second law of motion along -axis.

Rearrange the above expression for

From the free body diagram of rocket, the expression of net force along -axis is as follows, (there is only thrust and gravitational force along -axis)

Combine the above expression of net force along -axis and the expression of acceleration along -axis,

02

Step 2. Given.

Rearrange the derived expression of acceleration (in part a.) to find the mass of the rocket,

Further solve, to find the expression for mass,

Substitute for for and for in the above expression of mass, the mass of rocket remains is as follows,

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Most popular questions from this chapter

For an O2molecule, the constant 鈭�is approximately 0.00018eV. Estimate the rotational partition function for an O2molecule at room temperature.

In the low-temperature limit (kT<<鈭�), each term in the rotational partition function is much smaller than the one before. Since the first term is independent of T, cut off the sum after the second term and compute the average energy and the heat capacity in this approximation. Keep only the largest T-dependent term at each stage of the calculation. Is your result consistent with the third law of thermodynamics? Sketch the behavior of the heat capacity at all temperature, interpolating between the high-temperature and low- temperature expressions.

Use a computer to sum the exact rotational partition function numerically, and plot the result as a function ofkT鈭�. Keep enough terms in the sum to be confident that the series has converged. Show that the approximation in equation 6.31 is a bit low, and estimate by how much. Explain the discrepancy

The dissociation of molecular hydrogen into atomic hydrogen, H2鈫�2Hcan be treated as an ideal gas reaction using the techniques of Section 5.6. The equilibrium constant K for this reaction is defined as

K=PH2P0PH2

whereP0is a reference pressure conventionally taken to be1bar,and the other P's are the partial pressures of the two species at equilibrium. Now, using the methods of Boltzmann statistics developed in this chapter, you are ready to calculate K from first principles. Do so. That is, derive a formula for K in terms of more basic quantities such as the energy needed to dissociate one molecule (see Problem 1.53) and the internal partition function for molecular hydrogen. This internal partition function is a product of rotational and vibrational contributions, which you can estimate using the methods and data in Section 6.2. (AnH2 molecule doesn't have any electronic spin degeneracy, but an H atom does-the electron can be in two different spin states. Neglect electronic excited states, which are important only at very high temperatures. The degeneracy due to nuclear spin alignments cancels, but include it if you wish.) Calculate K numerically atT=300K,1000K,3000K,and6000K. Discuss the implications, working out a couple of numerical examples to show when hydrogen is mostly dissociated and when it is not.

For a COmolecule, the constant 鈧�is approximately 0.00024eV.(This number is measured using microwave spectroscopy, that is, by measuring the microwave frequencies needed to excite the molecules into higher rotational states.) Calculate the rotational partition function for a COmolecule at room temperature (300K), first using the exact formula 6.30 and then using the approximate formula 6.31
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