A 10 meter by 5 meter by 5 meter room is filled completely with Helium gas (near

Question

A 10 meter by 5 meter by 5 meter room is filled completely with Helium gas (nearly ideal). The mass of an individual atom can be calculated using the atomic mass number. For Helium this is 4u with u being the individual mass of a proton (or neutron) = 1.66x10-27 kg since Helium has 2 protons and 2 neutrons.

a. What is the mass of a single helium atom?

b.If the temperature of the room is lowered to 0°C, what is the rms velocity of the Helium atoms?

c.If the temperature of the room is to be increased to 50°C and the total mass of helium in the room is 50 kg, what amount of heat must be added?

Can you explain you answer? Thanks

Explanation / Answer

(a) The mass of the each helium atom is                    m   = 4 ( 1.66*10-27 kg)                          = 6.64*10-27 kg ------------------------------------------ ------------------------------------------- (b) The rms velocity of the helium atom is                     vrms = sqrt ( 3kT /m )                            = sqrt ( 3 (1.38*10-23 J /K)(273 K) / 6.64*10-27 kg )                                  = 1304 m/s -------------------------------------------------------- -------------------------------------------------------- (c) number of moles n = mass (m) / ( molar mass * avagadro number )                                 = ( ( 50 kg ) /   (( 6.64*10-27 kg ) *(6.023*1023 )                                 = 1.25*104 mol The heat energy added                            Q = nCv dT                                 = (1.25*104 mol )( 3/2 )(8.314 J/mol.K) ( 323 - 273K )                                 = 7.79*106 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---