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How Do You Calculate Kt/v

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At 200K, the particles of gas X have an average velocity equal to that of Argon at 400K. Identity of X ?
At 200K, the particles of gas X have an average velocity equal to that of Argon at 400K. What is the identity of X? HBr F2 CO HF He guessing its HBr...but im really not sure!! can somebody explain? thanks

KE(avg) = [(1/2)mv^2] = (3/2)kT v(rms) = √(3kT)/m = √(3RT)/M, where √ means square root, rms means root mean square, R = 8.3145 J/mol*K, T = absolute temperature. M = 39.948 g/mol(1 kg)/(1000 g) = 0.039948 kg/mol v(rms) = √[(3)(8.314 J/mol*K)(400 K)/0.039948 kg/mol] = 499.76 m/s This is the rms velocity of Argon. Now, substitute back into the equation for v(rms) and calculate for M, molecular mass. 499.76 m/s = √[(3)(8.314 J/mol*K)(200 K)/M] Square both sides and you get: 249760 m^2/s^2 = [(3)(8.314 J/mol*K)(200 K)/M] Now, solve for M: M = [(3)(8.314 J/mol*K)(200 K)/249760 m^2/s^2] = 0.019974 kg/mol Convert to g/mol: molar mass = 19.974 g/mol My best estimate is that the unknown gas HF. F = 18.998 g/mol; H = 1.0079 g/mol HF = 20.006 g/mol This is certainly good within experimental error.

Applying Binomial Theorem?
I'm trying to answer the following question 1) in the standard gas equation pV/T= k Volume increased by 3% Temperature diminished by 2% K remains constant. Find corresponding pressure change? This is all good but I have to do it using binomial theorem? I believe i have to get it in the form of (1+x)^n or for powers of small terms (1+nx) so i've re-arranged the equation to give me: p = kT(V^-1) but get abit lost from here on in? If anyone could shed any light on this I would be very grateful?

First step is to make the variable whose change you want to calculate the subject of the equation. In this case P = k.T/V We want to calculate the fractional or percentage change in P from the corresponding changes in the other variables, that is we want ΔP/P from ΔT/T and ΔV/V. The really simple way to derive the required relation is to take natural logs of the equation ln(P) = ln(k) + ln(T) - ln(V) and differentiate to get (since k is a constant) ΔP/P = ΔT/T - ΔV/V and there you have it - couldn't be quicker! However, we are required to involve the binomial theorem, so I guess we'd better derive the same equation the long way round. The actual changes in the variables in the ideal gas equation may be written P + ΔP = k.(T + ΔT).(V + ΔV)^(-1) Dividing the LHS by the LHS of the ideal gas equation in the initial state, and the RHS by the corresponding initial RHS yields (1 + ΔP/P) = (1 + ΔT/T).(1 + ΔV/V)^(-1) and now we expand each of the terms in brackets using the binomial expansion and noting that (1 + x)^n = 1 + n.x + ½.n.(n - 1).x² + .... and that if x is small, the third term may be neglected in comparison to the first two, if high accuracy is not required. For the same reason, terms involving products of small changes in different variables may also be ignored. 1 + ΔP/P = (1 + ΔT/T).(1 - ΔV/V) ~ 1 + ΔT/T - ΔV/V, or ΔP/P ~ ΔT/T - ΔV/V. So for ΔV/V = + 3% and ΔT/T = - 2%, ΔP/P = - 2 - 3 = - 5%.

Need fool solution on this please?
In penetrating through a plank of fluckness H, a bullet change its velocity from Vo(o stand 4 not)+V' calculate the time of motion of the bullet in the plank assuming the resistance force to be proportional to the cube of the velocity

ANSWER: T = ½H([Vo+V']^-1+Vo^-1). Write resistance force F=-mkV^3, where m is mass of bullet and k is a constant. Then acceleration of bullet in plank = resistance force/mass : dV/dt = -kV^3. Integrate between limits of (t=0, V=Vo) and (t,V): ∫ -V^-3.dV = ∫k.dt [½V^-2] = [kt] V^-2 - Vo^-2 = 2kt ...(1) . The bullet exits the plank when t=T and V=Ve=Vo+V', so Ve^-2 - Vo^-2 = 2kT ...(2). Rearrange (1) and integrate again wrt t : V = dX/dt = 1/√(2kt+Vo^-2) ∫dX = ∫1/√(2kt+Vo^-2).dt [X]= (1/k)[√(2kt+Vo^-2)] Apply limits (t=0, X=0) and (t=T, X=H) corresponding to the bullet entering and leaving the plank : H = (1/k)(√(2kT+Vo^-2)-√(2k.0+Vo^-2)) = (1/k)(√(2kT+Vo^-2) - Vo^-1)). Substitute for k using (2) and rearrange to find T : kH+Vo^-1 = √(2kT+Vo^-2) = √(Ve^-2-Vo^-2+Vo^-2) = Ve^-1 kH = Ve^-1-Vo^-1 ½(2kT)H = T(Ve^-1-Vo^-1) ½(Ve^-2 - Vo^-2)H = ½(Ve^-1 - Vo^-1)(Ve^-1+Vo^-1)H = T(Ve ^-1-Vo^-1). ANSWER: T = ½H(Ve ^-1+Vo^-1). Note: T = the average of the times taken to pass through the plank at the entry speed Vo and the exit speed Ve=Vo+V'. However, this is not generally true, it only applies for resistance proportional to cube of speed.

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