How Does Heat Affect the Expansion of Monatomic Gas in an Insulated Cylinder?

[alias25]

one mole of monatomic gas (in a perfectly insulated cylinder)
has temp = 27 degees c = 300K
volume = 0.025m^3

(gas held in frictionless piston)
heat suppied causes gas to expand at constant pressure gas expands to volume = 0.06m^3

i) calc final temp..
i used pV/T =pV/T (p cancels because its constant) and i got a value for new tempreture as 720K

ii) increase in av. KE per molecule...
using KE= 3/2kT i found the original KE per molecule as 1.49*10^-20
and new KE per molecule as 6.21*10^-21
and the difference between them (increase in av. KE = 8.69*10^-21)

iii) WD by the gas expanding...
WD = p*(change in)V...i got the p value from P = nRT/V and i got that as 99720 Nm-^2 and then multuply by (0.06 -0.025) to get WD as 3490.2 J

iiii) the heat supplied...
i used Q = (change in)u + (change in)WD
change in internal energy (u) is the sum of the change in KE of the molecules right?
so i multiplied change in av. KE per molecule 8.69*10^-21 by avagadros constant to give me KE for one mole, i get a value of 7284.2J

but then i thought that should be the same as...
calculating the change in av. KE of one mole so i used the formula KE = 3/2RT
to find change in KE for a mole and i get a value of 8974.8 - 3739.5 = 5235.3J

shouldnt they be the same, i don't know which ones wrong

anyway i used U as 7284.2J added (WD) 3490.2 to give me Q = 10774.4J

the other one would be Q = 5235.3 + 3490.2 =8725.5J
which ones right?! did i make any mistakes in the other questions?

 

[alias25]

don't worry i got it ...silly me 8.69*10^-21 multiplied by avagadros constant is not the value i got.

 

[quicknote]

Your calculations for the final temperature, increase in average kinetic energy, work done by the gas expanding, and heat supplied seem to be correct. However, there may be a small discrepancy between the two values for heat supplied because of rounding errors in your calculations.

To address your question about the difference in the two values for heat supplied, it is important to note that internal energy (u) is not just the sum of the change in kinetic energy of the gas molecules. It also includes the potential energy and internal energy of the gas molecules, which may change due to changes in temperature and volume. So, the formula Q = (change in)u + (change in)WD takes into account all of these changes in energy, whereas the formula Q = 3/2RT only considers the change in kinetic energy.

In this case, the difference in the two values for heat supplied is likely due to the change in internal energy of the gas molecules (which is not accounted for in the Q = 3/2RT formula). Therefore, the correct value for heat supplied is likely the one you calculated using the formula Q = (change in)u + (change in)WD.

Overall, your understanding and calculations of the first law of thermodynamics seem to be correct. Keep in mind that there may be small discrepancies in your calculations due to rounding errors. It is always a good idea to double check your calculations and make sure they are consistent with the principles of thermodynamics.