Asked by ~christina~
A house has well insulated walls. It contains a vol of 100m^3 of air at 300K
a) calculate the E required to increase the T of the diatomic gas by 1.00C
b) If this E could be used to lift an object of mass m through a height of 2.00m what is the mass of m?
a) calculate the E required to increase the T of the diatomic gas by 1.00C
b) If this E could be used to lift an object of mass m through a height of 2.00m what is the mass of m?
Answers
Answered by
drwls
a) Even thought the volume is known, this will not be a constant-volume heating of the gas. Some will expand through cracks around windows and doors, and through the chimney, if there is one. It is a constant-pressure heating, and the specific heat for a diatomic molecule in such a process is 7R/2. I don't have my calculagtor wth me, but that's about 7 Calories per mole degC. Use the perfect gas law and the volume to get the number of moles of gas in the house. Multiply by 4.18 if you want the energy in Joules.
b) Set the heat energy required (in J) equal to m g H, with H = 2 m. Solve for M.
b) Set the heat energy required (in J) equal to m g H, with H = 2 m. Solve for M.
Answered by
~christina~
Thanks very much drwls =)
Answered by
~christina~
I have a question about the pressure ..what do I use if it's not given?
Answered by
~christina~
do I just use 1atm ?
Answered by
drwls
Assume 1 atmosphere, about 1.02*10^5 N/m^2
Answered by
~christina~
Oh..I got a humongous number drwls.
I used PV=nRT
T= 274.15K
P= 1atm
V=100m^3
R= 8.2x10^-5 m^3* atm / K*mol
then I got 4,448.33mol
then I plugged that into
Q= nCp(DeltaT)
Cp= 7/2*R
T= 274.15K
n=4,448.33mol
Q= 35469438.74 J ?!
Why is it so bib
I used PV=nRT
T= 274.15K
P= 1atm
V=100m^3
R= 8.2x10^-5 m^3* atm / K*mol
then I got 4,448.33mol
then I plugged that into
Q= nCp(DeltaT)
Cp= 7/2*R
T= 274.15K
n=4,448.33mol
Q= 35469438.74 J ?!
Why is it so bib
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