November 15, 2019

Exam 1

Physics 208 – Introductory Physics II
Nazareth College
Department of Chemistry & Biochemistry
Robert F Szalapski, PhD – Adjunct Lecturer
www.CallMeDrRob.com
Spring 2012

Exam 1
Heat, Work, Ideal Gases and Phase Changes

§ Potentially Useful Equations


\displaystyle E \displaystyle= \displaystyle\frac{1}{2}mv^{2}
\displaystyle E \displaystyle= \displaystyle mgh
\displaystyle\Delta L \displaystyle= \displaystyle\alpha L_{0}\Delta T
\displaystyle\Delta V \displaystyle= \displaystyle\beta V_{0}\Delta T
\displaystyle PV \displaystyle= \displaystyle nRT
\displaystyle Q \displaystyle= \displaystyle m\, c\,\Delta T
\displaystyle Q \displaystyle= \displaystyle m\, L_{f}
\displaystyle Q \displaystyle= \displaystyle m\, L_{v}
\displaystyle T_{F} \displaystyle= \displaystyle\frac{9^{0}\mathrm{F}}{5^{0}\mathrm{C}}T_{C}+32^{0}\mathrm{F}
\displaystyle T_{C} \displaystyle= \displaystyle\frac{5^{0}\mathrm{C}}{9^{0}\mathrm{F}}\left(T_{F}-32^{0}\mathrm{F}\right)

§ Potentially Useful Data

Constants:

\displaystyle g \displaystyle= \displaystyle 9.80\frac{\mathrm{m}}{\mathrm{s}^{2}}
\displaystyle R \displaystyle= \displaystyle 8.314\frac{\mathrm{J}}{\mathrm{K}\cdot\mathrm{mole}}
\displaystyle= \displaystyle 1.99\frac{\mathrm{cal}}{\mathrm{K}\cdot\mathrm{mole}}
\displaystyle= \displaystyle 0.0821\frac{\mathrm{atm}\cdot\ell}{\mathrm{K}\cdot\mathrm{mole}}
\displaystyle N_{A} \displaystyle= \displaystyle 6.02\times 10^{{23}}\;\mathrm{mole}^{{-1}}

Conversion Factors:

\displaystyle 1\;\ell \displaystyle= \displaystyle 1000\;\mathrm{cm}^{3}\;\;\;(exact)
\displaystyle 1.000\;\mathrm{cm} \displaystyle= \displaystyle 0.3937\;\mathrm{in}
\displaystyle 1\;\mathrm{ft} \displaystyle= \displaystyle 12\;\mathrm{in}\;\;\;(exact)
\displaystyle 4.186\;\mathrm{J} \displaystyle= \displaystyle 1\;\mathrm{cal}

 

Substance c \left(\frac{\rm J}{\rm kg\cdot^{{0}}C}\right)
Ammonia (NH{}_{3}) 4,600
Water (l) 4,186
Ice 2,090
Ethanol 2,300
Steam 2,010
Beryllium (Be) 1,820
Air 1,004
Aluminum (Al) 900
Glass 837
Silicon (Si) 703
Iron (steel) 448
Copper (Cu) 387
Silver (Ag) 234
Gold (Au) 129
Lead (Pb) 128

 

Substance Fusion: L_{f}\left(\frac{\rm J}{\rm kg}\right) Vaporization: L_{v}\left(\frac{\rm J}{\rm kg}\right)
Water (H{}_{2}O) 33.5\times 10^{{4}} 22.6\times 10^{{5}}
Ammonia (NH{}_{3}) 33.2\times 10^{{4}} 13.7\times 10^{{5}}
Copper (Cu) 20.7\times 10^{{4}} 47.3\times 10^{{5}}
Benzene (C{}_{6}H{}_{6}) 12.6\times 10^{{4}} 3.94\times 10^{{5}}
Ethanol 10.8\times 10^{{4}} 8.55\times 10^{{5}}
Gold (Au) 6.28\times 10^{{4}} 17.2\times 10^{{5}}
Nitrogen (N{}_{2}) 2.57\times 10^{{4}} 2.00\times 10^{{5}}
Lead (Pb) 2.32\times 10^{{4}} 8.59\times 10^{{5}}
Oxygen (O{}_{2}) 1.39\times 10^{{4}} 2.13\times 10^{{5}}
Substance \alpha(\mbox{}^{0}\mathrm{C}^{{-1}}) \beta(\mbox{}^{0}\mathrm{C}^{{-1}})
Aluminum 25\times 10^{{-6}} 75\times 10^{{-6}}
Brass 19\times 10^{{-6}} 56\times 10^{{-6}}
Copper 17\times 10^{{-6}} 50\times 10^{{-6}}
Iron,Steel 12\times 10^{{-6}} 35\times 10^{{-6}}
Glass 9\times 10^{{-6}} 27\times 10^{{-6}}
Quartz 0.4\times 10^{{-6}} 1\times 10^{{-6}}
Gasoline 950\times 10^{{-6}}
Water 210\times 10^{{-6}}

§ Problem 1

On a cold Rochester morning when it is only 10^{0}\mathrm{F} outside you fill your automobile tank until it holds a total of 60\;\ell of gasoline. The temperature then climbs to 60^{0}\mathrm{F}. How large must your gasoline tank be at 60^{0}\mathrm{F} so that it does not overflow? (10 points)

First we must convert the temperature difference to Celsius.

\displaystyle\Delta T \displaystyle= \displaystyle\left(60^{0}\mathrm{F}-10^{0}\mathrm{F}\right)\frac{5^{0}\mathrm{C}}{9^{0}\mathrm{F}}
\displaystyle= \displaystyle\underline{2}7.8^{0}\mathrm{C}

Then

\displaystyle\Delta V \displaystyle= \displaystyle\beta V_{0}\Delta T
\displaystyle= \displaystyle\left(950\times 10^{{-6}}\mbox{}^{0}\mathrm{C}^{{-1}}\right)\left(60\;\ell\right)\left(\underline{2}7.8^{0}\mathrm{C}\right)
\displaystyle= \displaystyle\underline{6}2\;\ell

Note that most numbers were only specified to a single significant digit, so this is just an estimate. Nevertheless, it is clear that you do not want to fill your gasoline tank to the top of the fill pipe!

§ Problem 2

A 1.0\;\ell canister is open to the air at the room temperature of 0.0^{{0}}{\rm C}, and then the cover is closed. Assume that air is approximately an ideal gas.

§ (a)

How many moles of air are inside the canister? (10 points)

The easiest approach is to realize that we are at STP, and we know the molar volume for one mole. We could set up a simple proportion.

\displaystyle\frac{n}{1.0\;\ell} \displaystyle= \displaystyle\frac{1.00\;\mathrm{mole}}{22.4\;\ell}
\displaystyle n \displaystyle= \displaystyle 0.045\;\mathrm{mole}

The other approach is to use the Ideal Gas Law:

\displaystyle PV \displaystyle= \displaystyle nRT
\displaystyle n \displaystyle= \displaystyle\frac{PV}{RT}
\displaystyle= \displaystyle\frac{\left(1.00\;\mathrm{atm}\right)\left(1.0\;\ell\right)}{\left(0.0821\frac{\mathrm{atm}\cdot\ell}{\mathrm{K}\cdot\mathrm{mole}}\right)\left(273.15^{0}\mathrm{K}\right)}
\displaystyle= \displaystyle 0.045\;\mathrm{mole}

§ (b)

If the canister is heated to 75.0^{{0}}{\rm C} but the canister does not expand, then what is the new pressure inside the canister? (10 points)

\displaystyle\frac{P_{1}V}{P_{2}V} \displaystyle= \displaystyle\frac{nRT_{1}}{nRT_{2}}
\displaystyle P_{2} \displaystyle= \displaystyle\frac{P_{1}T_{2}}{T_{1}}
\displaystyle= \displaystyle\left(1.00\;\mathrm{atm}\right)\frac{273.15^{0}\mathrm{K}+75.0^{0}\mathrm{K}}{273.15^{0}\mathrm{K}}
\displaystyle= \displaystyle 1.\underline{2}8\;\mathrm{atm}
\displaystyle= \displaystyle 1.3\;\mathrm{atm}

§ (c)

Maintaining the temperature of 75.0^{{0}}{\rm C}, stomp on the canister until its volume is reduced by one-third. What is the pressure inside now? (10 points)

Reduced by one-third means

\displaystyle V_{2} \displaystyle= \displaystyle\frac{2}{3}\times 1.0\;\ell
\displaystyle\frac{P_{1}V_{1}}{P_{2}V_{2}} \displaystyle= \displaystyle\frac{nRT}{nRT}
\displaystyle P_{2} \displaystyle= \displaystyle P_{1}\frac{V_{1}}{V_{2}}
\displaystyle= \displaystyle\left(1.\underline{2}8\;\mathrm{atm}\right)\frac{1.0\;\ell}{\frac{2}{3}\times 1.0\;\ell}
\displaystyle= \displaystyle 1.9\;\mathrm{atm}

§ (d)

Allow the canister to return to the original room temperature. What is the final pressure? (10 points)

There are two good approaches. We can use the results from (c):

\displaystyle P_{2} \displaystyle= \displaystyle P_{1}\frac{T_{2}}{T_{1}}
\displaystyle= \displaystyle\left(1.\underline{2}8\;\mathrm{atm}\right)\frac{273.15^{0}\mathrm{K}}{273.15^{0}\mathrm{K}+75.0^{0}\mathrm{K}}
\displaystyle= \displaystyle 1.5\;\mathrm{atm}

The thing about the Ideal Gas Law is that it describes the current state of the gas in terms of the state variables. It is not important what path led to that state. Hence we could realize that only the pressure and volume are now different from the initial state while n and T are back to their original values. Hence

\displaystyle P_{2} \displaystyle= \displaystyle P_{1}\frac{V_{1}}{V_{2}}
\displaystyle= \displaystyle\left(1.0\;\mathrm{atm}\right)\frac{1.0\;\ell}{\frac{2}{3}\times 1.0\;\ell}
\displaystyle= \displaystyle 1.5\;\mathrm{atm}

§ Problem 3

100 g of \rm H_{2}O is initially at a temperature of -25^{{0}}{\rm C}. It is placed in a calorimeter where it is heated to 150^{{0}}{\rm C}. The temperature of the \rm H_{2}O is recorded as a function of time.

§ (a)

Make a plot of temperature versus time for the experiment. Label the axes, all temperatures of significance, and the phases involved in each segment of the graph. (10 points)

§ (b)

How much heat was required to heat the \rm H_{2}O? (Be sure to include heating and phase changes.) (20 points)

There are five steps, warming the ice, melting the ice to water, warming the water, boiling the water and heating the steam.

\displaystyle Q \displaystyle= \displaystyle mc_{\mathrm{ice}}\left[0^{0}\mathrm{C}-\left(-25^{0}\mathrm{C}\right)\right]+mL_{f}^{\mathrm{H_{2}O}}+mc_{\mathrm{water}}\left[100^{0}\mathrm{C}-0^{0}\mathrm{C}\right]
\displaystyle\mbox{}+mL_{v}^{\mathrm{H_{2}O}}+mc_{\mathrm{steam}}\left[150^{0}\mathrm{C}-100^{0}\mathrm{C}\right]
\displaystyle= \displaystyle\left(0.1\;\mathrm{kg}\right)\left\{\left(2090\frac{\mathrm{J}}{\mathrm{Kg}\cdot^{0}\mathrm{C}}\right)\left(25^{0}\mathrm{C}\right)+33.5\times 10^{4}\frac{\mathrm{J}}{\mathrm{Kg}}\right.
\displaystyle\mbox{}\left.+\left(4186\frac{\mathrm{J}}{\mathrm{Kg}\cdot^{0}\mathrm{C}}\right)\left(100^{0}\mathrm{C}\right)+22.6\times 10^{5}\frac{\mathrm{J}}{\mathrm{Kg}}+\left(2010\frac{\mathrm{J}}{\mathrm{Kg}\cdot^{0}\mathrm{C}}\right)\left(50^{0}\mathrm{C}\right)\right\}
\displaystyle= \displaystyle 320\;\mathrm{kJ}

§ (c)

Suppose the heat required all came from the 2 kg beryllium calorimeter which was initially at a temperature T_{i}^{\mathrm{Be}} but is in thermal equilibrium with the \rm H_{2}O at the end of the experiment. Calculate T_{i}^{\mathrm{Be}}. (10 points)

\displaystyle 0 \displaystyle= \displaystyle 320\;\mathrm{kJ}+m_{\mathrm{Be}}c_{\mathrm{Be}}\left(150^{0}\mathrm{C}-T_{i}^{\mathrm{Be}}\right)
\displaystyle T_{i}^{\mathrm{Be}} \displaystyle= \displaystyle\frac{320\;\mathrm{kJ}+m_{\mathrm{Be}}c_{\mathrm{Be}}\left(150^{0}\mathrm{C}\right)}{m_{\mathrm{Be}}c_{\mathrm{Be}}}
\displaystyle= \displaystyle\frac{320\;\mathrm{kJ}+\left(2\;\mathrm{kg}\right)\left(1820\frac{\mathrm{J}}{\mathrm{Kg}\cdot^{0}\mathrm{C}}\right)\left(150^{0}\mathrm{C}\right)}{\left(2\;\mathrm{kg}\right)\left(1820\frac{\mathrm{J}}{\mathrm{Kg}\cdot^{0}\mathrm{C}}\right)}
\displaystyle= \displaystyle 240^{0}\mathrm{C}

§ Problem 4

A student in Physics 208, wishing to test the equivalence of heat and energy, waits for a day when the temperature outside is exactly 0.0^{0}\mathrm{C}. An official men’s collegiate shot-put shot, mass m_{\mathrm{shot}}=7.26\;\mathrm{kg}, in thermal equilibrium with its surroundings, is then dropped a distance of 12\;\mathrm{m} into a pile of crushed ice. Estimate the mass m_{\mathrm{ice}} of ice that is melted as a result of stopping the shot. (10 points)

Gravitational potential energy will be converted to an equal amount of kinetic energy which in turn will be converted to an equal amount of heat which melts ice. We can leave out the middle step and equate gravitational potentially energy directly with heat.

\displaystyle m_{\mathrm{shot}}gh \displaystyle= \displaystyle Q
\displaystyle= \displaystyle m_{\mathrm{H_{2}O}}L_{f}^{\mathrm{H_{2}O}}
\displaystyle m_{\mathrm{H_{2}O}} \displaystyle= \displaystyle\frac{m_{\mathrm{shot}}gh}{L_{f}^{\mathrm{H_{2}O}}}
\displaystyle= \displaystyle\frac{(7.26\;\mathrm{kg})\left(9.8\frac{\mathrm{m}}{\mathrm{s}^{2}}\right)(12\;\mathrm{m})}{33.5\times 10^{4}\frac{\mathrm{J}}{\mathrm{Kg}}}
\displaystyle= \displaystyle 2.6\;\mathrm{g}

 

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