Question

This scenario describes questions 3-8: Remove Locking pin 50 cm 1.0 atm 20°C 100 cm 3.0 atm 1. Start 2. Heat to 3 atm. 3. Remove pin. Continue heating to 100 cm 4. Insert pin. Remove mass. The figure shows a sample of N2 gas covered by a movable piston, first in its initial state and then undergoing three processes. The heat capacity at constant volume for N2 is 5/2 R. In step 1, when the pin locking the piston is removed, the piston does not move. The piston has an area of 10 cm2. The figure gives the volume, pressure, and temperature of the gas. If there is a vacuum above the piston, what is the combined weight of the piston and masses, in Newtons? Please express all answers to three significant figures.

a) If there is a vacuum above the piston, what is the combined weight of the piston and masses, in Newtons?

b) How many moles of N₂ gas are in the sample?

c) (fill in the blank): In step 2, the pin is reinserted and the gas is heated until the pressure is 3.0 atmospheres. The new temperature and volume are __________ K and ________ cm³. The amount of work done on the gas is __________J; the amount of heat energy absorbed by the gas is ____________J, and the change in thermal energy of the gas is _________J. Remember, work and heat can both be either positive or negative.

d) (fill in the blank): In step 3, the pin is removed and the gas is allowed to expand. The new temperature and pressure are ________ K and ________ atm. The amount of work done on the gas is _________ J.

e) (select increases, decreases or stays the same for each of the following blanks)

In step 4, the pin is reinserted and one of the masses is removed. The pressure ____________, the volume  ________________ , and the temperature ____________ .

f) Draw a pV diagram for this series of processes.

Answer 1

Consider the area of pistor = A force on port piston by the masses of mi and ma socm² I atm 20°C F W=mig + m₂ g و كرا weight of piston = WAWI total force Pressure (P) uk A =10cm² = 10x10 4 m² P = 1 atm = 1.018105 N/m2 - 1.018105 F 10x10-4 - FF 1.01105x10x10-4 = 101 N. Combined weight of pisten and masses = 101 N. the formula PV = nRT (b using V = Pressure - P lata 1001x105 mm = 52 X10-6 m3 n = number of moles, R = 8.31 31mol-k. 50 cm T = 273 +20 = 29 3K PV an - RT 1.018105 X 50x10-6 (8.31) X 293 . ·002 mol. An Number of moles of N2 = .002 mol.

As pin locked the piston. the olime is constant. hele volume va sox10-6 m² Pressure-p= 3x1.010105 N/m² P. = 3 atm 2 .002 mole T = ? R = 8.31 Janie (mork using the formula, Pr=nRT. PV 3x101x105x50x10-6 *002 x (8.30 T = NR 911 - 55 K 912K. 639° e. a tento celsuus = 912-273 = temperature volume in Constant So work done W = P.Av. Ar= 0 W=o I once using the formula du tw [w=o - 8 =du en = > 2 change in internal energy = nC, dT. n = .002 mole Cv= 5R R=8-31 Joule/mol-k 2. dT= Change in temperature T2=912K T2 - Ti= 912 - 293 Ti = 293k 619 K. du = (1002) (5X8.31) x619 Jenke 25.7 Joule.

= heat energy a borbed by gas. du = 25.7 Joule volume = The new temperature 639°c. - 912K. motoru 5.00 xlo'cm? work done - 6. Change in internal energy = 25.7 Joul. heat absorbed = 25.7oule. d here pin is removed but temperature is same T=912K. 3 100cm T i Constant. This i isothermal process. PV = Constant Piri = P282 P = 3 atm. V = 50cm? V2=100cm3 P2=2 3 x 50 = P2X100 P2= 3XSO 1.5 atm wok done 2 in iso thermal prouss W = MRT in ( 13 ) n = .002 mol R= 8.313 1 molok, T= 9 12K N2 = 100cm? v=50cm? (1002) (8.31) (912) m (Leo) J 10.5 Jonke Am @ New temperature = 912k, Pressure = 1.50 atm work done = 10.5 J. 2 W=