{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ME320 solution4 - ME320 Thermodynamics Homework set#4 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
ME320 Thermodynamics Homework set #4 Due: Wednesday 10/1/03 at the start of class 1. 2–65C (Textbook) Solution: A gas can be treated as an ideal gas when it is at a high temperature or low pressure relative to the critical temperature and pressure. 2. 2–70 (Textbook) Solution: Initially, the absolute pressure in the tire is P 1 = P g + P atm = 210 kPa + 100 kPa = 310 kPa Treating the air as an ideal gas, it follows that P 1 V 1 T 1 = P 2 V 2 T 2 . Thus, P 2 = P 1 V 1 T 2 V 2 T 1 and, assuming that the volume of the tire remains constant, the final pressure in the tire is given by P 2 = P 1 T 2 T 1 = (310 kPa)(323 K) 298 K = 336 kPa Thus, the pressure rise is Δ P = P 2 - P 1 = (336 - 310) kPa = 26 kPa . Finally, using the gas constant of air given in Table A-1, we find that m 1 = P 1 V 1 RT 1 = (310 kPa)(0 . 025 m 3 ) (0 . 287 kPa m 3 / kg K)(298 K) = 0 . 0906 kg and m 2 = P 2 V 2 RT 2 = (310 kPa)(0 . 025 m 3 ) (0 . 287 kPa m 3 / kg K)(323 K) = 0 . 0906 kg , whereby the mass of air that needs to be bled off to restore pressure to its original value is Δ m = m 2 - m 1 = 0 . 0070 kg .
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
3. 2–76C (Textbook) Solution: The compressibility factor Z
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}