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