Stuck with physics gas questions

WoAiMeiMei

9 Apr 08, 20:07

Completed:
C19, P13 *after long confusion*
C19, P38
C19, P43

for P38 i i think you use Vavg and Vrms equations and rearrange to get m1/m2 ratio..... but im not quite sure how.

ahh!

newcomer

9 Apr 08, 20:18

damn i wish i listened to class on gases.

eagle

9 Apr 08, 22:09

i try out tmr night... got 1 jap oral test and 1 eng prof presentation tmr...

WoAiMeiMei

9 Apr 08, 22:16

cool.. these questions are giving me the sh-

>.<

^tamago^

10 Apr 08, 13:11

V₁ = 0.20m³
P₂ = 101.3 kPa
P₁ = 150 kPa + P₂
= 261.3 kPa

Since this is an isothermal process, ΔT = 0
Δ(PV) = ΔQ = nRΔT = 0
P₁V₁ = P₂V₂
V₂ = V₁P₁/P₂ = 0.20m³ × 261.3/101.3 = 0.516m³
W₁ = Work done by air = nRT ln(V_2/V₁) = P₁V₁ ln(V₂/V₁) = 0.20m³ × 261.3 kPa × ln(0.516m³/0.2m³) = 49.5kJ

During the cooling process, pressure is constant.
W₂ = Work done by air = p(V₁ - V₂) = 101.3kPa × (0.20m³ - 0.516m³) = -32.0kJ

∴ Total work done by air = W₁ + W₂ = 49.5kJ - 32.0kJ = 17.5kJ (3 s.f.)

WoAiMeiMei

10 Apr 08, 17:19

i cant get the right answer with 17509 Joules... argh!

weewee

10 Apr 08, 17:27

I dunno the formula but maybe tamago should use nRT In(V2-V1) instead of nRT ln(V2/V1).

WoAiMeiMei

10 Apr 08, 17:42

-

omg! wait i go recalc...

you applied the pressures to the volume calcs..

CRAP!

i got 17.5206 kJ

.. still wrong

ahhh!!! soo angry at this stupid work

^tamago^

10 Apr 08, 17:54

....

WoAiMeiMei

10 Apr 08, 17:56

V1 = 200L
P2 = 101.3 kPa
P1 = 261.3 kPa
V2 = 516L

W1 by air= nRT ln(V2/V1) = P1V1 ln(V2/V1) = 261.3 x 200 x ln (516/200)= 49531J
W2 on air = p(V2 - V1) = 101.3 Ã— (516-200)) = 32010.8

Total Work = W1 - W2 = 49513 - 32010.8
= 17502.2J

is still wrong.. TT

WoAiMeiMei

10 Apr 08, 18:15

Originally posted by ^tamago^:

....

editted.. just saw sth, be back soon

^tamago^

10 Apr 08, 18:17

yeah positive. what's ur model answer?

WoAiMeiMei

10 Apr 08, 18:21

heh... guess what my friend sent me

let me go through it first

yes!

it's correct >_<

ans = 15664.277 J

^tamago^

10 Apr 08, 18:21

Originally posted by weewee:

I dunno the formula but maybe tamago should use nRT In(V2-V1) instead of nRT ln(V2/V1).

nope.... cos u intergrate 1/V dV you will get ln v_f - ln v_i = ln(v_f/v_i).

^tamago^

10 Apr 08, 18:24

the variables look different. hmmm.

WoAiMeiMei

10 Apr 08, 18:28

yeah, the variables i use are randomly generated specially for me! yay !

the one in the solution is the one given in the original textbook question >_<

weewee

10 Apr 08, 19:27

Originally posted by ^tamago^:

nope.... cos u intergrate 1/V dV you will get ln v_f - ln v_i = ln(v_f/v_i).

Yeah as i state, i dunno the formula. TS's formula is wrong in #1 post.

Chapter 19, P13, i think you need to find the volume using the P1V1 = P2V2... then use the W = nRT ln (Vf - Vi)
which is also W = PV lv (Vf - Vi)

^tamago^

10 Apr 08, 19:34

Originally posted by weewee:

Yeah as i state, i dunno the formula. TS's formula is wrong in #1 post.

orh. paiseh. ><

weewee

10 Apr 08, 19:39

I am so ashamed to say i have never heard of this formula. Is this A level physics?

eagle

10 Apr 08, 20:24

Originally posted by weewee:

I am so ashamed to say i have never heard of this formula. Is this A level physics?

Uni Year 1 engineering physics...

btw, are all the questions solved already?

WoAiMeiMei

10 Apr 08, 20:24

it's umm...

... kinda >_<

the other questions are soooo argh

eagle

10 Apr 08, 20:27

I think I forgot all the formulas >_<

eagle

10 Apr 08, 20:37

C19 P38

temperature the same, kinetic energy of molecules should be the same

thus, (0.5)(m1)(v1) = (0.5)(m2)(v2)

but v2 = 2v1,

m1/m2 = v2/v1 = 2

Not confirmed though....

WoAiMeiMei

10 Apr 08, 20:39

nah 2 isnt the answer... there must be a reason why 5p is there

WoAiMeiMei

10 Apr 08, 20:41

looking at some formula...

p = (nMv^2)/3V , with v = root mean square speed

v(rms) = Sq root of (3RT/M)

V(avg) is top line of :