art A Classify the following by the sign of ?E for the system. Drag the appropri

Question

art A Classify the following by the sign of ?E for the system. Drag the appropriate items to their respective bins. If no definitive classification can be made, drag the item into the bin labeled

thank you , am at two tries at the moment :(

Part A Classify the following by the sign of AE for the system. Drag the appropriate items to their respective bins. If no definitive classification can be made, drag the item into the bin labeled "Not enough data." The system expands The system contracts The system expands and the surroundings and the surroundingsand the surroundings get hotter get hotter. get colder The system contracts and the surroundings get colder Negative Positive Not enough data

Explanation / Answer

a) The system expands There is a drop in temperature during expansion ( Excepto for hydrogen and helium). E of the system decreases, Surrounidings gets heated up and their. E increases.
b) The system contracts..E of the system increases and the surroundings get hotter..... E for the system decreases    There is "Not enough data"

c) The system expands..E for the system decreases and the surroundings get colder.... E for the system increases "Not enough data"

d)The system contracts. E for the system increases and the surroundings get colder.....E for the system increases "E for the system increases"

2. From First law of thermodynamics

E =Q+W

Q=-67.9 Kj (system releases heat)

W= -PdV= 30*(2.8-8)=-156atm.L

but 1 atm= 101.325 Joules, 156atm.L= 156*(101.325)=15807Joules= 15.81 Kj

E= -67.9+15.81=-52.09 Kj

3. for first stage E1= Q1+W1 = Q1+2*(4.4-2.2)*101.325 , E2= Q2+0.5*(2.2-1.76)*101.325

Q1 and Q2 are heat interactions for stage 1 and stage 2

for the overall process E = E1+ E2= Q1+Q2+267.498

Q is the heat interaction for overall

When change in brought in a single step E =Q+2.5*(4.4-1.76)*101.325= Q+283.71 Joules

Since final temperature remains the same and E is point function

Q1+Q2+267.498= Q+283.71

Q1+Q2-Q= 283.71-267.498=16.212 joules

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