HW11 - padilla(tp5647 – Homework 11 – Sutcliffe...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: padilla (tp5647) – Homework 11 – Sutcliffe – (53770) 1 This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The combustion of a certain amount of methane gives off 100 kJ of energy. Dur- ing this process the amount of energy in the universe 1. increases by 100 kJ. 2. The answer cannot be determined with- out knowing the heat capacity of methane. 3. decreases by 100 kJ. 4. does not increase or decrease. correct Explanation: The First Law of Thermodynamics states that the total amount of energy in the uni- verse is constant. The energy change occurs in the system. 002 10.0 points How much total work is done when 3 . 3 of H 2 expands from 25 liters to 50 liters against a constant external pressure of 1 atm, then ex- pands further from 50 to 100 liters against a constant external pressure of 0.1 atm? An- swer in calories. Correct answer:- 726. Explanation: w =- P Δ V w is not a state function. The pathway must be taken into consideration. Calculate w for each step. Step 1 : P = 1 atm V i = 25 L V f = 50 L Δ V = 25 L w = (- 1 atm)(25 L) parenleftbigg 101 . 33 J atm · L parenrightbigg × parenleftbigg 1 cal 4 . 1867 J parenrightbigg =- 605 . 0708 cal Step 2 : P = 0 . 1 atm V i = 50 L V f = 100 L Δ V = 50 L w =- (0 . 1 atm)(50 L) parenleftbigg 101 . 33 J atm · L parenrightbigg × parenleftbigg 1 cal 4 . 1867 J parenrightbigg =- 121 . 0142 cal summationdisplay w =- 605 . 0708 cal- 121 . 0142 cal =- 726 . 085 cal 003 10.0 points An ideal gas is allowed to expand isothermally from 2.00 liters at 5.00 atm to 5.00 liters at 2.00 atm against a vacuum ( i.e. , no outside pressure). Which of the quantities q , w , Δ E , and Δ H are zero for this process? 1. w and Δ E , 2. w , q , and Δ E 3. Some other combination 4. w 5. w , q , Δ E , and Δ H correct Explanation: P = 0 so w =- P Δ V = 0 . For isothermal expansion, T = const. There is no change in molecular kinetic en- ergy or potential energy due to intermolecu- lar attraction (which is zero for ideal gases). Therefore Δ E = 0. Δ E = q + w 0 = q + 0; q = 0 Δ H = q = 0 Therefore w , q , Δ E , and Δ H all equal zero. 004 10.0 points padilla (tp5647) – Homework 11 – Sutcliffe – (53770) 2 Which of the following is true of a general thermodynamic state function? 1. The change of the value of a state func- tion is independent of the path of a process. correct 2. The value of a state function does NOT change with a change in temperature of a process. 3. The change in the value of the state func- tion is always positive for endothermic pro- cesses. 4. The value of the state function remains constant....
View Full Document

Page1 / 10

HW11 - padilla(tp5647 – Homework 11 – Sutcliffe...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online