Question

NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Helium gas is compressed from 90 kPa and 30°C to 470 kPa in a reversible, adiabatic process. Determine the final temperature and the work done, assuming the process takes place in a piston-cylinder device. Use the table containing the ideal gas specific heats of various common gases. The final temperature is K. The work done is kJ/kg.

Answer 1

GIVEN: 1. Initial pressure = gokpa 2.) Initial temperature = 30°c Pressure = 470 kPa 4. Cras lived = Helium 3. Final Solution: IPARTI ideal gas from the table : Specific Heal ratio for Helium gas/v) = 1.667 Specific heat (CP) = 5.19 kJ/kg k Since the given process is reversible adiabatic process, so we knows relation : Tz P2 P 'Where T= temperatur tinel the Y- 1.667-1 Tz (470 l 7.667 Ti mitial temp P₂= final press Pia Initial pressen 90 (2 73430) =) T2 = 303 x 1.94 587 .04 K = 314°C Hence, final temperature is 587104k

v For piston-cylinder: Work clone = mali, -) he here, symbols Y-1 have their R (T2-T) r-1 2.0991 (5337.04-303) R for 1.667-1 Work clone per kg = heliem - 2.07 884.53 kolo (PART 2 in Steady hehen the process takes place flow Compressor: P.- gokkal P = 47okea Ti=300 =303k T = ? for temperature final, we will use the previous result so, T = 587.04k Work dones hq ha compressa = cp (t2-1) work done - haha) 5.19 (589-04-305) [For - 1474 17 kJ kJ/kg