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Chapter 19: Q22P (page 578)

Question: Find the rms speed of argon atoms at 313 K.Molar mass of argon atoms is 39.948 g/ mol.

Short Answer

Expert verified

Answer:

The RMS speed of argon atom is 442 m /s.

Step by step solution

01

Understanding the concept

We can find the RMS speed using its formula in terms of R, temperature, and molar mass.

The expression for the RMS speed is given by,

v=3RTM

02

Calculate the rms speed of argon atoms

The RMS speed of argon atoms is,

v=3RTM

Molar mass of argon isM=39.95脳10-3kg/mol.

v=38.31431339.95脳10-3v=442.058鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈥�鈮�442m/s

Therefore the RMS speed of argon atoms is 442 m /s .

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Most popular questions from this chapter

Question: Oxygen (O2) gas at 273 K and 1 atm is confined to a cubical container 10 cm on a side. Calculate螖鲍g/Kavg , where螖鲍g is the change in the gravitational potential energy of an oxygen molecule falling the height of the box and Kavg is the molecule鈥檚 average translational kinetic energy.

At273鈥�Kand1.00脳10-2atm, the density of a gas is1.24脳10-5g/cm3.

  1. Findvrmsfor the gas molecules.
  2. Find the molar mass of the gas
  3. Identify the gas.

We know that for an adiabatic process,pV纬=aconstant.Evaluate 鈥� constant鈥� for an adiabatic process involving exactly2.0鈥�molof an ideal gas passing through the state having exactlyp=1.0鈥�atmand T=300鈥�K. Assume a diatomic gas whose molecules rotate but do not oscillate

Suppose 1.00Lof a gas with纬=1.30, initially at 273K and1.00atm is suddenly compressed adiabatically to half its initial volume. Find

  1. Its final pressure
  2. Its final temperature
  3. If the gas is then cooled to 273Kat constant pressure, what is its final volume?

Two containers are at the same temperature. The first contains gas with pressurep1, molecular massm1, and RMS speedVrms1. The second contains gas with pressure2.0p1, molecular massm2, and average speedVavg2=2.0Vrms1. Find the mass ratiom1/m2.
See all solutions

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