If the volume of a gas is double at constant pressure, the average translational kinetic energy of its molecule will

CONCEPT:

• According to kinetic energy theory, if we increase the temperature of a gas, it will increase the average kinetic energy of the molecule, which will increase the motion of the molecules.
• This increased motion increases the outward pressure of the gas.
• The average kinetic energy of translation per molecules of the gas is related to temperature by the relationship:

\(\Rightarrow KE = \frac{3}{2}{k_B}T\)

Where E = kinetic energy, k_B = Boltzmann constant and T = temperature

CALCULATION:

Given V₂ = 2V₁ , P₂ = P₁

• The ideal gas law is given as,

\(\Rightarrow PV=nRT\)     -----(1)

Where P = pressure, V = volume, n = number of moles, R = gas constant and T = temperature in K

By equation 1,

\(\Rightarrow P_{1}V_{1}=nRT_{1}\)     -----(2)

\(\Rightarrow P_{2}V_{2}=nRT_{2}\)     -----(3)

By dividing equation 2 and equation 3,

\(\Rightarrow \frac{P_{1}V_{1}}{P_{2}V_{2}}=\frac{T_{1}}{T_{2}}\)

\(\Rightarrow \frac{PV_{1}}{P\times2V_{1}}=\frac{T_{1}}{T_{2}}\)

\(\Rightarrow T_{2}=2T_{1}\)     -----(4)

• The average kinetic energy for temperature T₁:

\(\Rightarrow KE_{1} = \frac{3}{2}{k_B}T_{1}\)     -----(5)

• The average kinetic energy for temperature T₂:

\(\Rightarrow KE_{2} = \frac{3}{2}{k_B}T_{2}\)

\(\Rightarrow KE_{2} = \frac{3}{2}{k_B}\times 2T_{1}\)

\(\Rightarrow KE_{2} =2\times \frac{3}{2}{k_B}T_{1}\)     -----(6)

By equation 5 and equation 6,

\(\Rightarrow KE_{2} =2KE_{1}\)

• Hence, option 1 is correct.