# Ch 14.1 Help

## Ch 14.1 Help

1) A gas expands at a constant pressure of 2.987 atm its volume from 9.813 liter to 19.99 liter. What is the work done by the gas (see sheet 7; include the correct sign) ?

2) 32.43 cal heat are added to a gas which does 211.2 J work by expanding. What is the change in internal energy of the gas (see sheet ? Indicate with a positive (negative) sign whether the internal energy increases (decreases).

3) 1.592 moles of a gas at room temperature (20 oC) expand quickly ("adiabatically") at an average pressure of 2.766 atm increasing its volume by 257.8 cm3. What is the final temperature in oC (see sheet 7,8,11') ? (NOTE! The change in volume given, not the volume itself. Thus you cannot use the gas equation to get the number of molecules.) Indicate with a positive (negative) sign whether the temperature decreases (increases).

4) The internal energy of a gas is quickly ("adiabatically") increased by 387.6 J at an average pressure of 4.257 atm. What is the change in volume (see sheet 7, ? Indicate with a negative (positive) sign whether the gas is compressed (expands).

__Answer__: 3079.3882086999997__How to solve__:2) 32.43 cal heat are added to a gas which does 211.2 J work by expanding. What is the change in internal energy of the gas (see sheet ? Indicate with a positive (negative) sign whether the internal energy increases (decreases).

__Answer__: -75.41558999999998__How to solve__:3) 1.592 moles of a gas at room temperature (20 oC) expand quickly ("adiabatically") at an average pressure of 2.766 atm increasing its volume by 257.8 cm3. What is the final temperature in oC (see sheet 7,8,11') ? (NOTE! The change in volume given, not the volume itself. Thus you cannot use the gas equation to get the number of molecules.) Indicate with a positive (negative) sign whether the temperature decreases (increases).

__Answer__: 16.359934588707677__How to solve__:4) The internal energy of a gas is quickly ("adiabatically") increased by 387.6 J at an average pressure of 4.257 atm. What is the change in volume (see sheet 7, ? Indicate with a negative (positive) sign whether the gas is compressed (expands).

__Answer__: -0.0008988157476414783__How to solve__:**periwinkle**- Posts : 22

Join date : 2008-09-17

## Questions 1&2

Question 1

W = P*Change in Volume

Change in volume = 10.177 (I just subtracted the two to find the difference) - keep in mind this is still in Liters though so divide that by 1000 to put into m^3 = .010177

Then I did # in atm * 101300 to put into pascals

2.987 * 101300 = 302583.1

Then i did 302583.1 * .010177 = 3079.388 answer is positive

Question 2

First, convert the calories to joules by multiplying by 4.187 since there are 4.187 J in 1 cal

32.43 cal * 4.187 = 135.78441

Then to find the difference in internal energy, subtract the other number you just calculated from the other number already given in joules

211.2 J - 135.78441 = 75.41559

Answer is negative

W = P*Change in Volume

Change in volume = 10.177 (I just subtracted the two to find the difference) - keep in mind this is still in Liters though so divide that by 1000 to put into m^3 = .010177

Then I did # in atm * 101300 to put into pascals

2.987 * 101300 = 302583.1

Then i did 302583.1 * .010177 = 3079.388 answer is positive

Question 2

First, convert the calories to joules by multiplying by 4.187 since there are 4.187 J in 1 cal

32.43 cal * 4.187 = 135.78441

Then to find the difference in internal energy, subtract the other number you just calculated from the other number already given in joules

211.2 J - 135.78441 = 75.41559

Answer is negative

**Kathleen**- Guest

## Question 4

Question 4

Q = DeltaU + W

They say that U(internal energy) = 387.6

W = P*Change in Volume

So you can plug in and do Q=U + P*Change in volume

It says on sheet 8 that for adiabatic Q=0 so do

0 = 387.6 + 431234.1*Change in volume

(I got the 431234.1 by multiplying the pressure in atm which was 4.257*101300)

You solve and get Change in Volume = -.0008988157

Q = DeltaU + W

They say that U(internal energy) = 387.6

W = P*Change in Volume

So you can plug in and do Q=U + P*Change in volume

It says on sheet 8 that for adiabatic Q=0 so do

0 = 387.6 + 431234.1*Change in volume

(I got the 431234.1 by multiplying the pressure in atm which was 4.257*101300)

You solve and get Change in Volume = -.0008988157

**Kathleen**- Guest

## Re: Ch 14.1 Help

I'm still working on question 3 but I assume you have to plug in both the equations into Q

So, where Q = DeltaU + W

You would do Q = (3/2)(n)(k)(T) + P(Change in Volume)

Possibly with a Tf-Ti where the T is and solve for Tf but that doesn't seem to be working for me, I converted all the values but I'm not getting anywhere yet. I'll let you know.

So, where Q = DeltaU + W

You would do Q = (3/2)(n)(k)(T) + P(Change in Volume)

Possibly with a Tf-Ti where the T is and solve for Tf but that doesn't seem to be working for me, I converted all the values but I'm not getting anywhere yet. I'll let you know.

**Kathleen**- Guest

## Question 3

Question 3

Delta U = -W

W = P*Change in Volume

P = 2.766 atm * 101300 = 280195.8 Pa

V = 257.8 cm^3 / 1000000 = .0002578 m^3

W = 280195.8 * .0002578 = 72.23447724

Delta U = (3/2)*N*k*(Tf-Ti)

N = 1.592 * (6.022e23) = 9.587024e23

K = constant which is 1.38e-23

Tf = final temperature, what you are looking to find

Ti = 20 degrees celsius

(3/2)*(9.587024e23)*(1.38e-23)*(Tf-20) = -72.234477

19.845139*Tf - 396.90279 = -72.234477

19.845139*Tf = 324.6683166

Tf = 16.36

Answer is positive

sometimes the second decimal will be off, but as long as you don't go past that you should be fine. I rounded to the third in my actual answer on maple TA and that worked so...idk

Delta U = -W

W = P*Change in Volume

P = 2.766 atm * 101300 = 280195.8 Pa

V = 257.8 cm^3 / 1000000 = .0002578 m^3

W = 280195.8 * .0002578 = 72.23447724

Delta U = (3/2)*N*k*(Tf-Ti)

N = 1.592 * (6.022e23) = 9.587024e23

K = constant which is 1.38e-23

Tf = final temperature, what you are looking to find

Ti = 20 degrees celsius

(3/2)*(9.587024e23)*(1.38e-23)*(Tf-20) = -72.234477

19.845139*Tf - 396.90279 = -72.234477

19.845139*Tf = 324.6683166

Tf = 16.36

Answer is positive

sometimes the second decimal will be off, but as long as you don't go past that you should be fine. I rounded to the third in my actual answer on maple TA and that worked so...idk

**Kathleen**- Guest

## in response to guest

(3/2)*(9.587024e23)*(1.38e-23)*(Tf-20) = -72.234477

19.845139*Tf - 396.90279 = -72.234477

19.845139*Tf = 324.6683166

~~~~~~~~~~~~~

First, i multiplied the (3/2)*(9.587024e23)*(1.38e-23) to get 19.84 then

I factored out with the parentheses so i multiplied the 19.84 by Tf and then by 20 so then I had 19.84Tf - 19.84*(20) = 19.84Tf - 396.90

19.845139*Tf - 396.90279 = -72.234477

19.845139*Tf = 324.6683166

~~~~~~~~~~~~~

First, i multiplied the (3/2)*(9.587024e23)*(1.38e-23) to get 19.84 then

I factored out with the parentheses so i multiplied the 19.84 by Tf and then by 20 so then I had 19.84Tf - 19.84*(20) = 19.84Tf - 396.90

**Kathleen**- Guest

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