Stuck with physics gas questions

last SG · 2008-04-10T20:41:05+00:00

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

eagle SG

I think I forgot all the formulas >_<

eagle SG

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 SG

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

WoAiMeiMei SG

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 :

WoAiMeiMei SG

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....

ahh i have a hint:x

You do not need all the information given. Did you write the equations for average speed and rms speed? How is molar mass related to molecular mass?

eagle SG

that's O level chemistry I think...

molar mass = molecular mass * mole number

eagle SG

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....

Oops... Careless with the above formula... Should be 1/2 m v^2

(0.5)(m1)(v1)^2 = (0.5)(m2)(v2)^2

but v2 = 2v1,

m1/m2 = (v2/v1)^2 = 4

WoAiMeiMei SG

i think that's related to the value "R"

ps: 4 isnt the ans

eagle SG

With this formula
p = (nMv^2)/3V

pV/n = Mv^2 / 3

this is a constant because R and T are constants here

so,

m1v1^2 / 3 = m2v2^2 / 3

Ratio of m1/m2 is still 4 leh...

I think different p means different V or n only... The Vrms is still heavily dependent on temperature.

WoAiMeiMei SG

arrr....

use Vavg and Vrms equations and rearrange to get m1/m2 ratio

thats how someone else got it >__< but how :O

eagle SG

OH!!!!!! SORRY.... didn't see it is Vavg!

According to your equations,

Vrms2 = Vavg * sqrt (3 * pi / 8) = 2.171 Vrms1

So should be 2.171^2 = 4.71?

WoAiMeiMei SG

4.71 is the correct answer! XD

eagle SG

apologize for being so careless :(

eagle SG

C19 P43

a) 0J as volume didn't change

change in T = 16.3
N = 2.42 mol
f = degree of freedom = 3 for monoatomic gas
k = boltzmann constant

change in thermal energy = 491.7 J => not confirmed

change in internal energy = a) + b) = 491.7J

d) should be 491.7J / (2.42 * 1mol) = 3.37e-22 J

WoAiMeiMei SG

thanks, that's the correct answer ^^

eagle SG

C19 P61

1.8 * 176^y = 5.51 * 79.2^y
2.2222^y = 3.061111
y = ln(3.061111) / ln(2.222222) = 1.4

=> diatomic gas

so,
310 * 176^1.4 = T * 79.2^1.4
T = 948.1K

c) PV = nRT
so n = PV/RT
p = 1.80atm, 1 atm = 101325 Pa
V = 176L = 176000cm3 = 0.176 m3
R = 8.314 J·K⁻¹mol^-1
T = 310K

thus, n = 6.92 mol

WoAiMeiMei SG

b and c cant get correct...hint is to apply the ideal gas law. anyways cant reply tonight..

eagle SG

C19 P61

1.8 * 176^y = 5.51 * 79.2^y
2.2222^y = 3.061111
y = ln(3.061111) / ln(2.222222) = 1.4

=> diatomic gas

so,
310 * 176^1.4 = T * 79.2^1.4
T = 948.1K

c) PV = nRT
so n = PV/RT
p = 1.80atm, 1 atm = 101325 Pa
V = 176L = 176000cm3 = 0.176 m3
R = 8.314 J·K⁻¹mol^-1
T = 310K

thus, n = 6.92 mol

Part b careless.... Equation is still correct... Anyway you should also help check for careless mistake mah... Not just take the values and use them without trying to understand how the equation was used...

The power should be 0.4, not 1.4

310 * 176^0.4 = T * 79.2^0.4
T = 426.65K

Part c is already ideal gas equation. Maybe you could help check if the calculations are wrong...

WoAiMeiMei SG

hmm im not sure for part c... i keep hitting a wall... will redo tonight.. still soo tired from last night

FirePig SG

« Previous Page Next Page »