Question

Two moles of an ideal gas are placed in a container whose volume is 6.4x 10 m the absolute pressure of the gas is 4.6 x 10 pa. What is the average translational kinetic energy of a molecule of the gas

Answer 1

Answer:

Use PV = nRT to find T.

Then use: (average KE) = (3/2)k

Answer 2

The average translation kinetic energy of a molecule of the gas is $366.65X10^{-23} Joules$.

Given,

Number of moles in the container=2

Volume of the container=6.4x10 $m^{3}$

Absolute pressure of the gas=4.6x10 Pascals.

To find,

the average translation kinetic energy of a molecule of the gas.

Solution:

• The kinetic energy of 1 mole of a gas can be expressed with the following equation:
• $K.E.=\frac{3}{2} PV$.
• where, P-pressure of the gas and V-volume of the gas.
• The kinetic energy of n moles of a gas can be expressed with the following equation:

• $K.E.=\frac{3}{2} nPV$.

• where, n-number of moles of gas.
• According to the ideal gas equation,
• $PV=nRT$
• R=8.314.
• The kinetic energy for n moles can be written as:
• $K.E.=\frac{3}{2}PV$.
• The kinetic energy of 1 molecule of a gas can be expressed with the following equation:
• $K.E.=\frac{3}{2}KT$.
• where, K-Boltzmann constant and T-absolute temperature(in Kelvin).
• $K=\frac{R}{Na}$.

The temperature of the gas will be,

$PV=nRT\\6.4X10X4.X10=2XRXT\\64X46=2XRXT\\T=\frac{64X46}{2XR} \\T=\frac{32X46}{R} \\T=\frac{1472}{R} Kelvin.$

The kinetic energy of 1 molecule of the gas will be,

$K.E.=\frac{3}{2}KT\\K.E.=\frac{3}{2}X\frac{R}{Na} X\frac{1472}{R} \\K.E.=\frac{3X736}{Na} \\K.E.=\frac{2208}{6.022X10^{23} }\\K.E.=366.65X10^{-23} Joules.$

Hence, the kinetic energy is $366.65X10^{-23} Joules$.

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