# Differences

This shows you the differences between two versions of the page.

 phy131studiof15:lectures:finalp2sol [2015/12/02 09:47]mdawber [Question 5] phy131studiof15:lectures:finalp2sol [2015/12/02 09:49]mdawber [Question 6] Both sides previous revision Previous revision 2015/12/02 09:49 mdawber [Question 6] 2015/12/02 09:49 mdawber [Question 6] 2015/12/02 09:47 mdawber [Question 5] 2015/12/02 09:46 mdawber [Question 4] 2015/12/02 09:43 mdawber [Question 3] 2015/12/02 09:42 mdawber [Question 2] 2015/12/02 09:40 mdawber [Question 2] 2015/12/02 09:39 mdawber [Question 1] 2015/11/30 14:52 mdawber created Next revision Previous revision 2015/12/02 09:49 mdawber [Question 6] 2015/12/02 09:49 mdawber [Question 6] 2015/12/02 09:47 mdawber [Question 5] 2015/12/02 09:46 mdawber [Question 4] 2015/12/02 09:43 mdawber [Question 3] 2015/12/02 09:42 mdawber [Question 2] 2015/12/02 09:40 mdawber [Question 2] 2015/12/02 09:39 mdawber [Question 1] 2015/11/30 14:52 mdawber created Line 180: Line 180: ===== Question 6 ===== ===== Question 6 ===== - a) For an adiabatic process + {{f11finalq7fig.png}} + + Santa slides down a chimney which has a volume of 10\,​m$^{3}$. The chimney is sealed at the bottom by a floor, and sealed at the top by Santa'​s belly. We can treat the air in the chimney as an ideal diatomic gas. (Note: Santa'​s suit has very good thermal insulation!) + + a) (5 points) When Santa slides down the chimney from top to bottom he reduces the volume of the air by a factor of 5 (ie. $V_{final}=\frac{V_{initial}}{5}=2\,​\mathrm{m^{3}}$). If the initial pressure is standard atmospheric pressure what is the final pressure. (Hint: No heat is added to the air in this process and we can approximate it is quasistatic.) + + For an adiabatic process $P_{i}V_{i}^{\gamma}=P_{f}V_{f}^{\gamma}$ where in this case $\gamma=\frac{7}{5}$ $P_{i}V_{i}^{\gamma}=P_{f}V_{f}^{\gamma}$ where in this case $\gamma=\frac{7}{5}$ Line 186: Line 192: $P_{f}=P_{i}(\frac{V_{i}}{V_{f}})^{\gamma}=101.3\mathrm{kPa}\times{5}^{7/​5}=964.2\mathrm{kPa}$ $P_{f}=P_{i}(\frac{V_{i}}{V_{f}})^{\gamma}=101.3\mathrm{kPa}\times{5}^{7/​5}=964.2\mathrm{kPa}$ - b) As the gas is ideal + b) (5 points) ​ If the initial temperature of the air in the chimney was $2\mathrm{^{o}C}$ what is the final temperature of the air in the chimney? + + As the gas is ideal $P_{i}V_{i}=nRT_{i}$ $P_{i}V_{i}=nRT_{i}$ Line 196: Line 204: $T_{f}=\frac{964.2}{101.3}\times\frac{1}{5}\times275=523.5\mathrm{K}$ $T_{f}=\frac{964.2}{101.3}\times\frac{1}{5}\times275=523.5\mathrm{K}$ - c) For an isothermal process + c) (5 points) To get Santa out of the chimney a fire is lit under it which adds heat to the air, so that it expands isothermally. How much work is done by the gas as it expands isothermally back to it's original volume. + + For an isothermal process $W=nRT\ln{\frac{V_{f}}{V_{i}}}$ $W=nRT\ln{\frac{V_{f}}{V_{i}}}$ Line 204: Line 214: $W=964.2\times10^{3}\times 2\ln(5)=3103\mathrm{kJ}$ $W=964.2\times10^{3}\times 2\ln(5)=3103\mathrm{kJ}$ - d) For an isothermal process + d) (5 points) What is the pressure in the chimney when Santa has returned to his original position? + + For an isothermal process $P_{i}V_{i}=P_{f}V_{f}$ $P_{i}V_{i}=P_{f}V_{f}$ Line 210: Line 222: $P_{f}=\frac{V_{f}}{V_{i}}P_{i}=\frac{964.2}{5}=192.8\mathrm{kPa}$ $P_{f}=\frac{V_{f}}{V_{i}}P_{i}=\frac{964.2}{5}=192.8\mathrm{kPa}$ - e) Red is first process, green is second + e) (5 points) Sketch a pressure-volume diagram which shows both processes, making sure to label each process and the direction in which they occur. + + Red is first process, green is second {{finalf11q7solfig.png}} {{finalf11q7solfig.png}}