A beam of hydrogen molecules H2 is directed toward a wall, atn angle of 55o the

Question

A beam of hydrogen molecules H2 is directed toward a wall, atn angle of 55o the normal to the wall. Each molecule in the beam has a speedof 1.0 km/s and a mass of 3.3 x 10^-24 g. The beam strikes the wallover an area of 2.0 cm^2, at the rate of 10^23 molecules persecond. What is the beam's pressure on the wall?
Formulas of concern: nM(vrms^2) / 3V where M represents the molar mass of Hydrogen which is givenin my textbook as 2.02 x 10^-3 kg/mol
also the textbook give Vrms = 1920 m/s for hydrogen molecule @room temperature
n (number of moles in a sample) = Msample/M
I calculated n
n= 3.3x10^-27 kg / 2.02 x 10 ^-3 kg/mol n=1.634 x 10 ^-24 mol
Then i got extremely stuck as I am unsure how to getVolume..Could anyone help me please?
I know and (Vavg)^2=3p/ where represents density, p representspressure

Explanation / Answer

   Given :   A = 2.0 cm2 = 2.0 x 10-4m2       m = 3.3 x10-27 kg,   
     N / t = 1023moles/s,   
      v = 1.0 x 103m/s

The change in momentum is       p = Ft
   (2Nmvcos) = Ft     F = 2(N/t)mvcos     Pressure = F/A
                  =(2(N/t)mvcos)/A                  = (2 * 1023 moles/s * 3.3 x 10-27 kg *1.0 x103 m/s * cos55o)/(2.0 x 10-4m2)                   =1.89 kPa
I hope it helps you       m = 3.3 x10-27 kg,   
     N / t = 1023moles/s,   
      v = 1.0 x 103m/s

The change in momentum is       p = Ft
   (2Nmvcos) = Ft     F = 2(N/t)mvcos     F = 2(N/t)mvcos     Pressure = F/A
                  =(2(N/t)mvcos)/A                  = (2 * 1023 moles/s * 3.3 x 10-27 kg *1.0 x103 m/s * cos55o)/(2.0 x 10-4m2)                   =1.89 kPa

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