# How Do You Calculate Kt/v

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Question
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 ..im 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.

Question
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?