. Advertisement .

..3..

. Advertisement .

..4..

......... ADVERTISEMENT .........

..8..

......... ADVERTISEMENT .........

..8..

......... ADVERTISEMENT .........

..8..

**Part A**Determine the pressure (in atm) of the gas at points 1, 2, and 3

_{}

*p*= 0.999,5.0,0.999 atm

_{1},p_{2},p_{3 }**Part B**Determine the temperature (in °C) of the gas at points 1, 2, and 3

*T*= °C

_{1},T_{2},T_{3}**Part C**Determine the volume (in cm

^{3}) of the gas at points 1, 2, and 3.

*V*= cm

_{1},V_{2},V_{3}^{3}

**Part D**How much work is done on the gas during each of the three

*W*= J

_{1→2},W_{2→3},W_{3→1 }**Part E**How much heat energy is transferred to or from the gas during each of the three segments?

*Q*= J

_{1→2},Q_{2→3},Q_{3→1}*ΔU=Q−W*

m = given mass = 110 mg = 0.110 g

M = molar mass = 4 g/mol

N = Number of Moles

n = M/M = 0.110/4 =0.0275

Part A

Point 1:

T

_{1}= Temperature At Point 1 = 133C = 133 + 273 = 406 KV

_{1}= Volume At Point 1 = 1000 cm^{}= 10^{ -3}MP

_{1}= Pressure at Point 1 =?Use the equation

P_{1}V_{1}= n RT

_{1}Inserting the values

P

_{1}(10^{-3}) = (0.0275) (8.314) (406)P= 92825.8 p =_{1}0.916atm

Point 2:

P= 5 P_{2}_{1}= 5 (0.916) =4.58 atmPoint 3:

P= P_{3}_{1}=0.916atm

Part B)

T_{1}= 133^{o}CProcess 1 and 2:

P(Since volume is constant)_{1}/T_{1}=P

_{2}/T_{2}P

_{1}/(133 + 273) = 5P_{1}/T_{2}T

_{2}= 2030KT= 2030 – 273 =_{2}1757

^{o}CT

_{3}= T_{2}= 1757^{o}CPart C

V_{1}= 1000 cm^{3}V= V_{2}_{1}=1000cm

^{3}V

_{1}/T_{1}= V_{3}/T_{3}1000/(133 + 273) = V

_{3}/2030V_{3}= 5000 cm^{3}Part D

for process 1-2 :V = 0

W= P_{1-2}_{1}V =0for process 2-3 :W

_{2-3}= N R T_{2}log (V_{3}/V_{2}).W= (0.0275) (8.314) (2030) log(5000/1000) =_{2-3}747 Jfor process 3-1:WP_{3-1}=_{1}(V_{1}– V_{3}) = (92825.8) (0.001 – 0.005) =– 371.3J