August 1, 2021

Solution Set 1

Physics 208-02 – Introductory Physics II
Solution Set 1
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
Fall 2011

§ Chapter 13, Problem 5

§ (a)

15^{0} below zero on the Celsius scale is what Fahrenheit temperature?

\displaystyle T_{F} \displaystyle= \displaystyle\frac{9}{5}\frac{{}^{0}\mathrm{F}}{{}^{0}\mathrm{C}}T_{C}+32^{0}\mathrm{F}
\displaystyle= \displaystyle\frac{9}{5}\frac{{}^{0}\mathrm{F}}{{}^{0}\mathrm{C}}(-15^{0}\mathrm{C})+32^{0}\mathrm{F}
\displaystyle= \displaystyle 5^{0}\mathrm{F}

§ (b)

15^{0} below zero on the Fahrenheit scale is what Celsius temperature?

\displaystyle T_{C} \displaystyle= \displaystyle\frac{5}{9}\frac{{}^{0}\mathrm{C}}{{}^{0}\mathrm{F}}\left(T_{F}-32^{0}\mathrm{F}\right)
\displaystyle= \displaystyle\frac{5}{9}\frac{{}^{0}\mathrm{C}}{{}^{0}\mathrm{F}}\left(-15^{0}\mathrm{F}-32^{0}\mathrm{F}\right)
\displaystyle= \displaystyle-26^{0}\mathrm{C}

§ Chapter 13, Problem 7

Figure 1.

Chapter 13, Problem 7.

A concrete highway is built of slabs 12\;\mathrm{m} long at 20^{0}\mathrm{C}. How wide should the expansion cracks be (at 20^{0}\mathrm{C}) to prevent buckling if the range of temperatures is -30^{0}\mathrm{C} to +50^{0}\mathrm{C}?

We will need the coefficient of linear expansion for concrete given by

\displaystyle\alpha \displaystyle\approx \displaystyle 12\times 10^{{-6}}\left({}^{0}\mathrm{C}\right)^{{-1}}\;.

The coefficient is positive, so we should expect that the concrete will contract at lower temperatures and expand at higher temperatures. The maximum expansion will be by an amount

\displaystyle\Delta L \displaystyle= \displaystyle\alpha L\Delta T
\displaystyle= \displaystyle\left(12\times 10^{{-6}}\left({}^{0}\mathrm{C}\right)^{{-1}}\right)\left(12\;\mathrm{m}\right)\left(50^{0}\mathrm{C}-20^{0}\mathrm{C}\right)
\displaystyle= \displaystyle 4.3\times 10^{{-3}}\mathrm{m}
\displaystyle= \displaystyle 4.3\;\mathrm{mm}\;.

As we see in the figure, if we assume that the slab expands symmetrically, then with an expansion by \Delta L, it expands by \frac{1}{2}\Delta L into each gap, but the adjacent slab does the same. Therefore the gap must be \Delta L.

The gaps between the slabs are often filled with a compressible board that sets the gap properly. This is because the gaps may fill with water, and water has the unusual property that it actually expands when it freezes. Hence, allowing the gaps to fill with ice can lead to buckling and fracturing.

§ Chapter 13, Problem 10

To make a secure fit rivets that are larger than the rivet hole are often used and the rivet is cooled (usually in dry ice) before it is placed in the hole. A steel rivet 1.871\;\mathrm{cm} in diameter is to be placed in a hole 1.869\;\mathrm{cm} in diameter at 20^{0}\mathrm{C}. To what temperature must the rivet be cooled to fit into the hole?

We will need the coefficient of linear expansion for steel given by

\displaystyle\alpha \displaystyle= \displaystyle 12\times 10^{{-6}}\left({}^{0}\mathrm{C}\right)^{{-1}}\;.


\displaystyle\Delta L \displaystyle= \displaystyle\alpha L\Delta T

may be rewritten as

\displaystyle\Delta T \displaystyle= \displaystyle\frac{\Delta L}{\alpha L}
\displaystyle= \displaystyle\frac{\left(1.869\;\mathrm{cm}-1.871\;\mathrm{cm}\right)}{\left(12\times 10^{{-6}}\left({}^{0}\mathrm{C}\right)^{{-1}}\right)\left(1.871\;\mathrm{cm}\right)}
\displaystyle= \displaystyle-89^{0}\mathrm{C}\;.

Indeed the negative sign indicates cooling.

\displaystyle\Delta T \displaystyle= \displaystyle T_{f}-T_{i}
\displaystyle T_{f} \displaystyle= \displaystyle\Delta T+T_{i}
\displaystyle= \displaystyle-89^{0}\mathrm{C}+20^{0}\mathrm{C}
\displaystyle= \displaystyle-69^{0}\mathrm{C}\;.

At atmospheric pressure dry ice has a temperature of -78.5^{0}\mathrm{C} or colder, so it would be adequate as suggested by the problem statement.

§ Chapter 13, Problem 12

A quartz sphere is 8.75\;\mathrm{cm} in diameter. What will be its change in volume if it is heated from 30^{0}\mathrm{C} to 200^{0}\mathrm{C}?

We will need to know the initial volume of the sphere.

\displaystyle V_{0} \displaystyle= \displaystyle\frac{4}{3}\pi r^{3}
\displaystyle= \displaystyle\frac{4}{3}\pi\left(\frac{8.75\;\mathrm{cm}}{2}\right)^{3}
\displaystyle= \displaystyle 35\underline{0}.8\;\mathrm{cm}^{3}


\displaystyle\Delta V \displaystyle= \displaystyle\left(1\times 10^{{-6}}\;\;\mbox{}^{0}\mathrm{C}^{{-1}}\right)\left(35\underline{0}.8\;\mathrm{cm}^{3}\right)\left(200^{0}\mathrm{C}-30^{0}\mathrm{C}\right)
\displaystyle= \displaystyle 0.060\;\mathrm{cm}^{3}

§ Chapter 13, Problem 13

An ordinary glass is filled to the brim with 350.0\;\mathrm{ml} of water at 100.0^{0}\mathrm{C}. If the temperature decreased to 20.0^{0}\mathrm{C}, how much water could be added to the glass?

Calculate the change in volume for the glass and for the water.

\displaystyle\Delta V_{{H_{2}O}} \displaystyle= \displaystyle\left(210\times 10^{{-6}}\;\;\mbox{}^{0}\mathrm{C}^{{-1}}\right)\left(350.0\;\mathrm{ml}\right)\left(20.0^{0}\mathrm{C}-100.0^{0}\mathrm{C}\right)
\displaystyle= \displaystyle-5.88\;\mathrm{ml}\;,
\displaystyle\Delta V_{{glass}} \displaystyle= \displaystyle\left(27\times 10^{{-6}}\;\;\mbox{}^{0}\mathrm{C}^{{-1}}\right)\left(350.0\;\mathrm{ml}\right)\left(20.0^{0}\mathrm{C}-100.0^{0}\mathrm{C}\right)
\displaystyle= \displaystyle-0.76\;\mathrm{ml}\;.

To calculate the available space, subtract the volume of the water from the volume of the glass.

\displaystyle V_{{glass}}-V_{{H_{2}O}} \displaystyle= \displaystyle\left(350.0\;\mathrm{ml}+\Delta V_{{glass}}\right)-\left(350.0\;\mathrm{ml}+\Delta V_{{H_{2}O}}\right)
\displaystyle= \displaystyle\Delta V_{{glass}}-\Delta V_{{H_{2}O}}
\displaystyle= \displaystyle\left(-0.76\;\mathrm{ml}\right)-\left(-5.88\;\mathrm{ml}\right)
\displaystyle= \displaystyle 5.12\;\mathrm{ml}

§ Chapter 13, Problem 29

If 3.00\;\mathrm{m}^{3} of a gas initially at STP is placed under a pressure of 3.20\;\mathrm{atm}, the temperature of the gas rises to 38.0^{0}\mathrm{C}. What is the volume?

Pressure, volume and temperature are all changing. Only the quantity measured in moles remains the same.

\displaystyle P_{1}V_{1} \displaystyle= \displaystyle nRT_{1}
\displaystyle P_{2}V_{2} \displaystyle= \displaystyle nRT_{2}

Dividing the first equation by the second,

\displaystyle\frac{P_{1}V_{1}}{P_{2}V_{2}} \displaystyle= \displaystyle\frac{T_{1}}{T_{2}}
\displaystyle V_{2} \displaystyle= \displaystyle\frac{P_{1}V_{1}T_{2}}{P_{2}T_{1}}

Using STP to determine the initial temperature and pressure,

\displaystyle V_{2} \displaystyle= \displaystyle\frac{(1.00\;\mathrm{atm})(3.00\;\mathrm{m}^{3})(273.15^{0}\mathrm{K}+38.0^{0}\mathrm{K})}{(3.20\;\mathrm{atm})(273.15^{0}\mathrm{K})}
\displaystyle= \displaystyle 1.07\;\mathrm{m}^{3}

§ Chapter 13, Problem 30

In an internal combustion engine air at atmospheric pressure and a temperature of about 20^{0}\mathrm{C} is compressed in the cylinder by a piston to about \frac{1}{9} of its original volume (compression ratio=9.0). Estimate the temperature of the compressed air assuming the pressure reaches 40\;\mathrm{atm}.

This is similar to the previous problem, so we begin with

\displaystyle\frac{P_{1}V_{1}}{P_{2}V_{2}} \displaystyle= \displaystyle\frac{T_{1}}{T_{2}}

We are also given

\displaystyle V_{2} \displaystyle= \displaystyle\frac{1}{9}V_{1}
\displaystyle V_{1} \displaystyle= \displaystyle 9V_{2}

which simplifies the equation further.

\displaystyle\frac{P_{1}\; 9V_{2}}{P_{2}V_{2}} \displaystyle= \displaystyle\frac{T_{1}}{T_{2}}
\displaystyle\frac{P_{1}\; 9}{P_{2}} \displaystyle= \displaystyle\frac{T_{1}}{T_{2}}
\displaystyle T_{2} \displaystyle= \displaystyle\frac{T_{1}P_{2}}{9P_{1}}

Substituting the numbers,

\displaystyle T_{2} \displaystyle= \displaystyle\frac{(273^{0}\mathrm{K}+20^{0}\mathrm{K})(40\;\mathrm{atm})}{9(1\;\mathrm{atm})}
\displaystyle= \displaystyle\underline{1}300^{0}\mathrm{K}
\displaystyle= \displaystyle 1000^{0}\mathrm{C}

§ Chapter 13, Problem 33

A storage tank at STP contains 18.5\;\mathrm{kg} of nitrogen (\mathrm{N}_{2}).

§ (a)

What is the volume of the tank?

Here we must use the full form of the Ideal Gas Law.

\displaystyle PV \displaystyle= \displaystyle nRT
\displaystyle V \displaystyle= \displaystyle\frac{nRT}{P}

First we must find the number of moles of \mathrm{N}_{2} gas noting that the molecular weight is twice the atomic weight of nitrogen.

\displaystyle n \displaystyle= \displaystyle\frac{(18.5\;\mathrm{kg})\frac{1000\;\mathrm{g}}{1\;\mathrm{kg}}}{2\left(14.0\frac{\mathrm{g}}{\mathrm{mole}}\right)}
\displaystyle= \displaystyle 661\;\mathrm{mole}


\displaystyle V \displaystyle= \displaystyle\frac{nRT}{P}
\displaystyle= \displaystyle\frac{(661\;\mathrm{mole})\left(0.0821\frac{\ell\mathrm{atm}}{\mathrm{mole}^{0}\mathrm{K}}\right)\left(273^{0}\mathrm{K}\right)}{1.00\;\mathrm{atm}}
\displaystyle= \displaystyle 14,800\ell
\displaystyle= \displaystyle(661\;\mathrm{mole})(22.4\ell)

The last statement is a good check of our math since we know the molar volume of a mole of ideal gas at STP.

§ (b)

What is the pressure if an additional 15.0\;\mathrm{kg} of nitrogen is added without changing the temperature?

Allowing only n and P to change,

\displaystyle P_{1}V \displaystyle= \displaystyle n_{1}RT
\displaystyle P_{2}V \displaystyle= \displaystyle n_{2}RT

Dividing the first equation by the second,

\displaystyle\frac{P_{1}}{P_{2}} \displaystyle= \displaystyle\frac{n_{1}}{n_{2}}
\displaystyle P_{2} \displaystyle= \displaystyle\frac{P_{1}n_{2}}{n_{1}}

We now have

\displaystyle n_{2} \displaystyle= \displaystyle\frac{(18.5\;\mathrm{kg}+15.0\;\mathrm{kg})\frac{1000\;\mathrm{g}}{1\;\mathrm{kg}}}{2\left(14.0\frac{\mathrm{g}}{\mathrm{mole}}\right)}
\displaystyle= \displaystyle 12\underline{0}0\;\mathrm{mole}


\displaystyle P_{2} \displaystyle= \displaystyle\frac{(1.00\;\mathrm{atm})(12\underline{0}0\;\mathrm{mole})}{661\;\mathrm{mole}}
\displaystyle= \displaystyle 1.81\;\mathrm{atm}


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