Chemical Eng: How do I get value M from 2 unknown variables.

[Luke Attigan]

Homework Statement

At T1 the root mean square speed (rms) of compound P is 485.2 m s-1 and at T2 the rms speed is 504.1 m s-1. Using this data and the fact that T2 – T1 = 24.0 °C, determine the molar mass of P in g mol-1.

Homework Equations

1. From problem statement you have: (3·R·T1/M)1/2 = 485.2 and (3·R·T2/M)1/2 = 504.1
2. You also have that T2 - T1 = 24.0 °C ≡ 24.0 K
3. Together, (1) and (2) form a system of 3 algebraic equations with 3 unknowns (T1, T2 and M) that can be easily solved
4. Solve the equation system by first calculating T1 and T2
5. Once T1 and T2 are known use one of the expression in (1) to calculate M

The Attempt at a Solution

I have the 3 equations:485.2 = (3RT1/M)^0.5

504.1 = (3RT2/M)^0.5

T2 – T1 = 24 K

I have squared these equations, (rms)^2 = 3RT/M, in order to eliminate the square root to make the equation as simple as possible, however that is all I have so far and do not understand how to calculate M with two three unknown variables T1, T2 and M. Could you please give me some assistance?

 

[DrClaude]

Since you have $T_2 - T_1$ in the third equation, try to get a $T_2 - T_1$ from the other (squared) equations.

 

[SteamKing]

Homework Statement

At T1 the root mean square speed (rms) of compound P is 485.2 m s-1 and at T2 the rms speed is 504.1 m s-1. Using this data and the fact that T2 – T1 = 24.0 °C, determine the molar mass of P in g mol-1.

Homework Equations

1. From problem statement you have: (3·R·T1/M)1/2 = 485.2 and (3·R·T2/M)1/2 = 504.1
2. You also have that T2 - T1 = 24.0 °C ≡ 24.0 K
3. Together, (1) and (2) form a system of 3 algebraic equations with 3 unknowns (T1, T2 and M) that can be easily solved
4. Solve the equation system by first calculating T1 and T2
5. Once T1 and T2 are known use one of the expression in (1) to calculate M

The Attempt at a Solution

I have the 3 equations:485.2 = (3RT1/M)^0.5

504.1 = (3RT2/M)^0.5

T2 – T1 = 24 K

I have squared these equations, (rms)^2 = 3RT/M, in order to eliminate the square root to make the equation as simple as possible, however that is all I have so far and do not understand how to calculate M with two three unknown variables T1, T2 and M. Could you please give me some assistance?

Find the ratio of T2 : T1 by dividing the appropriate equation by the other. All of the other quantities like R and M will cancel.

Once you know the ratio of T2:T1, you can express T2 in terms of T1, and since you know the difference T2 - T1 = 24, you can solve for T1.

 

[Luke Attigan]

Since you have $T_2 - T_1$ in the third equation, try to get a $T_2 - T_1$ from the other (squared) equations.

Hi DrClaude. I don't know if I have did this correctly, but my logic is this:

I have rearranged (3RT1/M)^0.5 to become (rms)^2 = 3RT/M

What I have now done is this: (rms)^2 = 3RT2/M
T2 = [M.(rms)^2] \ 3R
__________________ =
T1 = [M.(rms)^2] \ 3R

Is this correct?

 

[DrClaude]

Your equation is coming out funny (not sure what the = sing in the middle is doing there). But now that you have equations for T1 and T2, you can either get T2-T1 as I suggested, or T2/T1 as SteamKing suggested.

 

[Luke Attigan]

Your equation is coming out funny (not sure what the = sing in the middle is doing there). But now that you have equations for T1 and T2, you can either get T2-T1 as I suggested, or T2/T1 as SteamKing suggested.

Hi again!

I think you both have helped me out enough now. I greatly appreciate it.

What I've did is T2 = [M.(rms)^2] \ 3R] / T1 = [M.(rms)^2] \ 3R]

(T1+24)/T1 = ratio

Solve for T1

Then use that figure to solve for T2.

I'm hoping this is correct.

I hope you both will be able to help me in the future - I'm finding some of the course very hard and quite horrible.

Luke.