a) How much work is done on the gas as the temperature of 0.155 mol of the gas is raised from 30.0° to 340°C?
______ J
b) What does the sign of your answer to part (a) indicate?
- the gas does positive work on its surroundings
- there is no work done, by the gas or the surroundings
- The surroundings do positive work on the gas
2. Gas in a container is at a pressure of 1.8 atm and the volume of 9.0 m3.
a) What is the work done on the gas if it expands at constant pressure to twice its initial volume?
_____ J
b) What is the work done on the gas if it is compressed at constant pressure to one quarter of its initial volume?
______ J
3.A gas expands from I to F along the three paths indicated in the figure below. Calculate the work done on the gas along each of the following paths.
a) IAF
______ J
b) IF
______ J
c) IBF
______ J
4. One mole of an ideal gas initially at a temperature of Ti = 1.8°C undergoes an expansion at a constant pressure of 1.00 atm to three times its original volume
a) Calculate the new temperature Tf of the gas.
______ K
b) Calculate the work done on the gas during the expansion.
_____ kJ
5. One mole of an ideal gas initially at a temperature of 1.50 x 102°C is compressed at a constant pressure of 2.00 atm to two-thirds its initial volume.
a) What is the final temperature of the gas?
_____ K
b) Calculate the work done on the gas during the compression.
_____ J
6. An ideal gas in a cylinder is compressed very slowly to one-third its original volume while its temperature is held constant. The work required to accomplish this task is 64 J.
a) What is the change in the internal energy of the gas?
_____ J
b) How much energy is transferred to the gas by heat in this process?
_____ J
7. A gas is compressed at a constant pressure of 0.800 atm from 9.00 L to 2.00 L. In the process, 390 J of energy leaves the gas by heat.
a) What is the work done on the gas?
_____ J
b) What is the change in its internal energy?
_____ J
8. Five moles of gas initially at a pressure of 2.00 atm and a volume of 0.300 L has internal energy equal to 82.0 J. In its final state, the gas is at a pressure of 1.50 atm and a volume of 0.800 L, and its internal energy equals 168 J.
a) For the paths IAF, IBF, and IF in the figure above, calculate the work done on the gas.
WIAF = ______ J
WIBF = ______ J
WIF = ______ J
b) For the paths IAF, IBF, and IF in the figure above, calculate the net energy transferred to the gas by heat in the process.
QIAF = _____ J
QIBF = _____ J
QIF = _____ J
9. A heat engine operates between a reservoir at 24°C and one at 439°C. What is the maximum efficiency possible for this engine?
______
10.In each cycle of its operation, a heat engine expels 3,000 J of energy and performs 1,700 J of mechanical work.
a) How much thermal energy must be added to the engine in each cycle
______ J
b) Find the thermal efficiency of the engine.
_____ %
11. One of the most efficient engines ever built is a coal-fired steam turbine engine in the ohio valley, driving an electric generator as it operates between 1870°C and 430°C.
a) What is its maximum theoretical efficiency?
______ %
b) Its actual efficiency is 42.0%. How much mechanical power does the engine deliver if it absorbs 1.8 x 105 J of energy each second from the hot reservoir?
_____ kW
12. An engine absorbs 1.68 kJ from a hot reservoir at 277°C and expels 1.16 kJ to a cold reservoir at 27°C in each cycle.
a) What is the engine's efficiency?
______ %
b) How much work is done by the engine in each cycle?
______ J
c) What is the power output of the engine if each cycle lasts 0.305 s?
______ kW
13. A heat pump has a coefficient of performance of 3.75 and operates with a power consumption of 6.90 x103 W.
a) How much energy does it deliver into a home during 6 h of continuous operation?
______ J
b) How much energy does it extract from the outside air?
______ J
14. A heat engine operates in a Carnot cycle between 82.0°C and 340°C. It absorbs 21,600 J of energy per cycle from the hot reservoir. The duration of each cycle is 4.00 s.
a) What is the mechanical power output of this engine?
_____ kW
b) How much energy does it expel in each cycle by heat?
______ kJ
Hey Mike, it's not accepting your math on #2 b. It is giving the message: "Use the expression relating the pressure and change in volume to the work done."
ReplyDeleteI'll take a look at it and see what the problem is. Can you send me the numbers you are working with (I'm under the impression that webassign changes the red numbers in every question so that nobody gets the same question).
DeleteLMK what your numbers are and I'll try to troubleshoot and see what is wrong... whether its on my end or maybe calculator trouble (IDK).
I just re-watched #2. It's definitely worked correctly. It may be that you are plugging numbers in your calculator weird.
DeleteIf you let me know what your specific numbers are and what answer your getting that Web Assign doesn't like, then I'll be able to find out what is wrong.
Hi Mike, I would like to ask why question#7, the work done on gas is positive instead of negative.
ReplyDeleteA Ching, whenever a gas system expands it is doing work on the environment, whenever it shrinks, the environment is doing work on the gas.
DeleteTo put it another way, we are trying to calculate how much energy was added to the gas by compressing it. In thermodynamics, work is a form of mechanical energy, so how much extra energy was added to the gas by work.
Does that answer your question?
I was wondering if you could help me out with this problem, Webassign has changed a lot so a lot of the questions are different. Thanks
ReplyDeleteA quantity of a monatomic ideal gas undergoes a process in which both its pressure and volume are increased by a factor of n = 9 as shown in the figure below. What is the energy absorbed by heat into the gas during this process? Hint: The internal energy of a monatomic ideal gas at pressure P and occupying volume V is given by
U = 3/2PV.
(Give your answer as a multiple of P0V0.)
https://www.webassign.net/sercp9/12-p-016-alt.gif
The first thing is to look at what your given and ask how it relates to what your asked for:
ReplyDelete1. Your given the change of P and the change of V-- and it changed such that the area under the PV curve could always be calculated by Multiplting P and V
2. Your also given an equation to calculate internal energy -- and therefore, the change in internal energy...
3. You're asked for the transfer of heat energy (the change of heat energy)... not the same thing as internal energy.
The change of internal energy (the energy that a system has in all of its forms)
Delta U (change of internal energy) is equal to all of the energy that comes in minus all of the energy that leaves.
--in this case I believe the two things that are important are:
--Work done by the system (the energy of work)
--heat transferred into the system.
Delta U = Delta heat - delta work
You have an equation for internal energy
U=3/2PV
So DeltaU must be
3/2PV(final state) - 3/2PV(initial state)
Work can be calculated -- in an ideal gas, work is always the area under the PV curve (it's always a square for you)
W = PV
--so the change of work is:
PV(final state) - PV(initial state)
Then we can combine all of this together--first, we solve our internal energy equation for "heat"
Delta U = Q - W
(Q is the heat transfer)
Q = deltaU+W
Then our substation
Q = [3/2PV(final) - 3/2PV(initial)] + [PV(final) - PV(initial)]
ONLY ONE THING LEFT
- we have to define PV(final)
The problem said that both increased by a factor of "n" (nP) (nV)
So PV(final) = nPnV = n^2PV (n-squared multiplied by PV)
Now substitute in our PV (final)
Change of internal energy
Delta U= 3/2 n^2 PV - 3/2PV
-- simplify by factoring our 3/2PV
Delta U = 3/2PV(n^2 -1)
Work done
n^2PV - PV
Simplify by factoring
PV(n^2 - 1)
Substitute everything into Q=DeltaU + W
Q = 3/2PV(n^2 - 1) + PV(n^2 - 1)
Simplify:
Q = 5/2PV(n^2 - 1)
Your question stated that n=9
9^2 - 1 = 80
Q = 5/2 X 80 X PV
Q = 200 X PV
So, if I didn't miss anything, your question wanted you to give the final answer as a multiple of PV--the answer is 200
- if you're wondering how I got 5/2, it's because PV(n^2 - 1) has a coefficient of 1
--- 1PV(n^2 - 1) ---this was my equation for work, so ...algebra!
Hope that helps... lmk if I missed anything. I'm doing this on the fly--on my phone in a parking garage, so, not writing anything down or checking any books...I'm about 90% confident.