Question-8 : The kinetic energy of a gas at temperature 90 degree celsius, which is given as -
using an equation, we have
K.E = (3/2) k T
where, k = 1.38 x 10-23 J/K
T = temperature = 90 0C = 363 K
then, we have
K.E = (3/2) (1.38 x 10-23 J/K) (363 K)
K.E = 751.4 x 10-23 J
K.E = 7.51 x 10-21 J
Question-10 : Her initial moment of inertia was given as -
using conservation of angular momentum,
I11 = I22
I1 (9.42 rad/s) = (0.88 kg.m2) (25.1 rad/s)
I1 = (22.08 kg.m2 rad/s) / (9.42 rad/s)
I1 = 2.35 kg.m2
please show a solution Determine the kinetic energy of a gas at temperature 90 degree C,...
A cylinder of volume 1.00m^3 contains 320 g of oxygen gas (O_2) at a temperature of 27.0 degree C. The atomic mass of oxygen is 16 u. (1u = 1.66 times 10^-27 kg) What is the pressure exerted by the gas on the walls of the container? If the volume of the cylinder is reduced to half its original value by pushing down on a piston and the pressure exerted by the oxygen gas on the walls of the cylinder...
Part C What is the rms speed of an oxygen gas molecule at room temperature (20.0°c)? Atomic mass of oxygen gas molecule - 32.0 u 1 atomic mass uni 1.66 x 10-27 kg m/s
Part C What is the rms speed of an oxygen gas molecule at room temperature (20.0°c)? Atomic mass of oxygen gas molecule - 32.0 u 1 atomic mass uni 1.66 x 10-27 kg m/s
PartC What is the rms speed of an oxygen gas molecule at room temperature (20.0°C)? Atomic mass of oxygen gas molecule- 32.0 u 1 atomic mass unit 1.66 x 1027 kg 四? m/s Submit Request Answer ▼ Part D
QUESTION 16 According to the kinetic theory of gases, the average kinetic energy of the gas particles in a gas sample is directly proportional to the O pressure. O volume O temperature O molar mass O number of moles of gas
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