Question

Two moles of an ideal gas are placed in a container whose volume is 2.3 x 10^-3 m3. The absolute pressure of the gas is 6.9 x 10^5 Pa.
What is the average translational kinetic energy of a molecule of the gas?

Answer 1

Answer:

$K.E.=1.97\times 10^{-21}\ J$

Explanation:

Given that:-

Pressure = $6.9\times 10^5\ Pa$

The expression for the conversion of pressure in Pascal to pressure in atm is shown below:

P (Pa) = $\frac {1}{101325}$ P (atm)

Given the value of pressure = 43,836 Pa

So,  

$6.9\times 10^5\ Pa$ = $\frac{6.9\times 10^5}{101325}$ atm

Pressure = 6.80977 atm

Volume = $2.3\times 10^{-3}\ m^3$ = 2.3 L ( 1 m³ = 1000 L)

n = 2 mol

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

6.80977 atm × 2.3 L = 2 mol × 0.0821 L.atm/K.mol × T

⇒T = 95.39 K

The expression for the kinetic energy is:-

$K.E.=\frac{3}{2}\times K\times T$

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

T is the temperature

So, $K.E.=\frac{3}{2}\times 1.38\times 10^{-23}\times 95.39\ J$

$K.E.=1.97\times 10^{-21}\ J$