Answer:
The work done on neon = -323 J
The internal energy change= -392.84 J
The heat absorbed by neon = -69.84 J
Explanation:
Step 1: Data given
Number of moles = 0.500 moles
Pressure = 1 atm
Temperature = 273 Kelvin
The pressure will change from 1.00 atm to 0.200 atm. The temperature changes from 273 to 210 Kelvin.
a) calculate the work done on neon
W = -P(V2-V1)
⇒ with P = the pressure = 0.1 atm
⇒ with V1 = the initial volume = nRTi /Pi
⇒ with V2 = the final volume = nRTf /Pf
W = -PnR((T2/P2) -(T1/P1))
⇒ with T2 = the final temperature = 210 K
⇒ with T1 = the initial temperature = 273 K
⇒ with P2 = the final pressure = 0.200 atm
⇒ with P1 = the initial pressure = 1.00 atm
W = -nR (210*(0.1/0.2) - 273*(0.1/1.00))
W = -nR*(105 - 27.3)
W= -(0.500)*(8.314)*(77.7)
W = -323 J
b) calculate the internal energy change
E = (3/2)*nRT
ΔE = Ef - Ei
ΔE =(3/2)*nR(T2-T1)
⇒ with n= number of moles = 0.500 moles
⇒ with T2 =the final temperature = 210 K
⇒ with T1 = the initial temperature = 273 K
ΔE = (3/2)*(0.5)*(8.314)(210-273)
ΔE = -392.84 J
c) Calculate the heat absorbed by neon
ΔE = q + W
q = ΔE -W
⇒ with ΔE = -392.84 J
⇒ with W = -323 J
q = -392.84 J -( -323 J)
q =-392.84 J + 323 J
q = -69.84 J
<span>1 trial : you have nothing to compare the result with - you don't know if it's a mistake.
2 trials : you can compare results - if very different, one may have gone wrong, but which one?
3 trials : if 2 results are close and 3rd far away, 3rd probably unreliable and can be rejected.
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First calculate the enthalpy of fusion. M, C and m,c = mass and
specific heat of calorimeter and water; n, L = mass and heat of fusion
of ice; T = temperature fall.
L = (mc+MC)T/n.
c=4.18 J/gK. I assume calorimeter was copper, so C=0.385 J/gK.
1. M = 409g, m = 45g. T = 22c, n = 14g
L = (45*4.18+409*0.385)*22/14 = 543.0 J/g.
2. M = 409g, m = 49g, T = 20c, n = 13g
L = (49*4.18+409*0.385)*20/13 = 557.4 J/g.
3. M = 409g, m = 54g, T = 20c, n = 14g
L = (54*4.18+409*0.385)*20/14 = 547.4 J/g.
(i) Estimate error in L from spread of 3 results.
Average L = 549.3 J/g.
average of squared differences (variance) = (6.236^2+8.095^2+1.859^2)/3 = 35.96
standard deviation = 5.9964
standard error = SD/(N-1) = 5.9964/2 = 3 J/g approx.
% error = 3/547 x 100% = 0.5%.
(ii) Estimate error in L from accuracy of measurements:
error in masses = +/-0.5g
error in T = +/-0.5c
For Trial 3
M = 409g, error = 0.5g
m = 463-409, error = sqrt(0.5^2+0.5^2) = 0.5*sqrt(2)
n =(516-463)-(448-409)=14, error = 0.5*sqrt(4) = 1.0g
K = (mc+MC)=383, error = sqrt[2*(0.5*4.18)^2+(0.5*0.385)^2] = 2.962
L = K*T/n
% errors are
K: 3/383 x 100% = 0.77
T: 0.5/20 x 100% = 2.5
n: 1.0/14 x 100% = 7.14
% errors in K and T are << error in n, so we can ignore them.
% error in L = same as in n = 7% x 547.4 = 40 (always round final error to 1 sig fig).
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The result is (i) L= 549 +/- 3 J/g or (ii) L = 550 +/- 40 J/g.
Both are very far above accepted figure of 334 J/g, so there is at least
one systematic error in the experiment or the calculations.
eg calorimeter may not be copper, so C is not 0.385 J/gK. (If it was
polystyrene, which absorbs/ transmits little heat, the effective value
of C would be very low, reducing L.)
Using +/- 40 is probably best (more cautious).
However, the spread in the actual results is much smaller; try to explain this discrepancy - eg
* measurements were "fiddled" to get better results; other Trials were made but only best 3 were chosen.
* measurements were more accurate than I assumed (eg masses to nearest 0.1g but rounded to 1g when written down).
Other sources of error:
L=(mc+MC)T/n is too high, so n (ice melted) may be too small, or T (temp fall) too high - why?
* it is suspicious that all final temperatures were 0c - was this
actually measured or just guessed? a higher final temp would reduce L.
* we have assumed initial and final temperature of ice was 0c, it may
actually have been colder, so less ice would melt - this could explain
small values of n
* some water might have been left in container when unmelted ice was
weighed (eg clinging to ice) - again this could explain small n;
* poor insulation - heat gained from surroundings, melting more ice,
increasing n - but this would reduce measured L below 334 J/g not
increase it.
* calorimeter still cold from last trial when next one started, not
given time to reach same temperature as water - this would reduce n.
Hope This Helps :)
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The phase of matter where atoms lose their electrons is plasma. It requires quite a bit of heat, too.
Answer:
A neutral solution hope it helps
