You have a bottle containg a mole of a monatomic gas such as hellum or neon. You

Question

You have a bottle containg a mole of a monatomic gas such as hellum or neon. You warm monatomic gas with an electrical heater which inputs Q=510 joules of thermal energy transfer. How much does the temperature of the gas increase? deltaT=0.000349 K You have a bottle containing a mole of a diatomic gas such as nitrogen(N_2) or oxygen (O_2). The initial temperature is the range where many rotational energy levels are excited but essentially no vibrational energy levels. You warm up this diatomic gas with an electrical heater which inputs Q=510 joules of thermal energy transfer. How much does the temperature of the gas increase? deltaT= K You have a bottle containing a mole of a diatomic gas such as nitrogen(N_2) or oxygen (O_2). The initial temperature is in the range where not only many rotational energy levels are excited but also many vibrational energy levels are excited. You warm up this diatomic gas with an electrical heater which inputs Q=510 joules of thermal energy tr

Explanation / Answer

You use the equations from the lecture where for monatomic it's:(3/2)kT, diatomic without vibrational: (5/2)kT, and with vibrational: (7/2)kT
Also you include Avogadro's number because it's per atom.

Boltzmann’s constant k = 1.4*10-23 J/K

Avogadro's number Na = 6.02*1023 mol-1

Part A
(3/2)(1.4*10-23 J/K)(6.02*1023 mol-1)(T) = 510 J
then T = 510 / (3/2)(1.4*10-23  J/K)(6.02*1023 mol-1)

T = 40.34 K

Part B
(5/2)(1.4*10-23  J/K)(6.02*1023 mol-1)(T) = 510

then T = 510 / (5/2)(1.4*10-23  J/K)(6.02*1023 mol-1)

T = 24.20 K

Part C
(7/2)(1.4*10-23)(6.02*1023 mol-1)(T) = 510
then T = 510 / (7/2)(1.4*10-23  J/K)(6.02*1023 mol-1)

T = 17.28 K

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