Suppose 0.021 mole of an ideal monatomic gas has a pressure of 1.00 atm at 273.15 K and is then cooled at constant pressure until its volume is reduced by 41.1 %. What is the amount of work done by the gas?

Answer 1

Use the gas law and solve for the volume.

PV=nRT

Where
P=1.00 atm
V=?
R=0.0821 liter·atm/mol·K
T=273.15K
and
n=0.021

V=nRT

Answer should contain 2-significant figures, but keep 3 for intermediate calculations.

V*0.411=Final Volume

Final Volume-Volume calculated from gas law=Change in volume.

Change in volume*1.00 atm=Work

Answer 2

Use the gas law and solve for the volume.

PV=nRT

Where
P=1.00 atm
V=?
R=0.0821 liter·atm/mol·K
T=273.15K
and
n=0.021

*** Typo correction

V=nRT/P

Answer should contain 2-significant figures, but keep 3 for intermediate calculations.

V*0.411=Final Volume

Final Volume-Volume calculated from gas law=Change in volume.

Change in volume*1.00 atm=Work

Answer 3

I converted from 1 atm to Pa, then I found answear 28107.55 J, but answear not correct..

Answer 4

To get work done in joules, use metres, Pa, and m³ as units.

Volume at 0° and 1 atm for n=0.021
V= 22.4*0.021 L
= 4.704*10^-4

Pressure 1 atm
= 101.3 kPa
= 101300 Pa

ΔV
=0.41V
=1.929 *10^-4

Using
Work=PΔV,
=101300*1.929*10^-4
= 19.54 J
= 20 J (approx.)

Answer 5

Oops, the reduction in volume is 41.1%, so ΔV=1.933 *10^-4 m³

and work done,
W=22.4*0.021/1000*0.411*101300
=19.58 j
=20 j (approx.)

Answer 6

You can use the fact that 1 mole of a gas occupy's a volume of 22.4L at STP, which gives you the initial volume faster then using the gas law, but both ways will return the same initial volume.

I suspect that you made an error in your calculations, and the work value for the system should be negative.

****You can just convert atm*L at the end, or do it the way that Mathmate indicated; either way should return the correct answer.

V=nRT/P

V=(0.021 moles)*(0.0821 liter·atm/mol·K)*(273.15K)/1.00atm

V=0.47L

Vf=0.471L*0.411

1 atm*L= 101.325 J

0.194L-0.4704L=-0.2775L

1.00 atm(-0.276L)=0.276 atm*L

0.276 atm*L*(101.325 J/1 atm*L)=-28 J

Answer 7

** Fixed a typo: look for **

You can use the fact that 1 mole of a gas occupy's a volume of 22.4L at STP, which gives you the initial volume faster then using the gas law, but both ways will return the same initial volume.

I suspect that you made an error in your calculations, and the work value for the system should be negative.

****You can just convert atm*L at the end, or do it the way that Mathmate indicated; either way should return the correct answer.

V=nRT/P

V=(0.021 moles)*(0.0821 liter·atm/mol·K)*(273.15K)/1.00atm

V=0.47L

Vf=0.471L*0.411=0.194L**

1 atm*L= 101.325 J

0.194L-0.4704L=-0.2775L

1.00 atm(-0.276L)=0.276 atm*L

0.276 atm*L*(101.325 J/1 atm*L)= -28 J (The answer is negative)