Question 608: Thermodynamics & KTG

Question

Select the correct statement.

Accepted Answer

Checking Translational K.E.:
The formula for translational kinetic energy is $	ext{Translational K.E.} = \frac{3}{2} \mathrm{PV}.$
This means that options (A) and (D) are wrong, because they do not match this formula.
Finding r.m.s. Speed:
The r.m.s. speed ($\mathrm{v}_{\mathrm{rms}}$) is given by $\mathrm{v}_{\mathrm{rms}} = \sqrt{\frac{3RT}{\mathrm{M}}}.$
This shows that r.m.s. speed increases as the square root of the temperature: $\mathrm{v}_{\mathrm{rms}} \propto \sqrt{\mathrm{T}}.$
Checking Option (C):
For option (C): $\mathrm{T}_1 = 100^\circ \mathrm{C}$, $\mathrm{T}_2 = 627^\circ \mathrm{C}$
First, convert to Kelvin:
$\mathrm{T}_1 = 100 + 273 = 373\,\mathrm{K}$
$\mathrm{T}_2 = 627 + 273 = 900\,\mathrm{K}$
Calculate the ratio of speeds:
$ \frac{\mathrm{v}_1}{\mathrm{v}_2} = \sqrt{\frac{\mathrm{T}_1}{\mathrm{T}_2}} = \sqrt{\frac{373}{900}} = \frac{\sqrt{373}}{30} $
This means:
$ \mathrm{v}_2 = \frac{30}{\sqrt{373}}\,\mathrm{v}_1 $
So, option (C) is not correct.
Checking Option (B):
For option (B): $\mathrm{T}_1 = 73^\circ \mathrm{C}$, $\mathrm{T}_2 = 527^\circ \mathrm{C}$
Convert to Kelvin:
$\mathrm{T}_1 = 73 + 273 = 346\,\mathrm{K}$
$\mathrm{T}_2 = 527 + 273 = 800\,\mathrm{K}$
Calculate the ratio of speeds:
$ \frac{\mathrm{v}_1}{\mathrm{v}_2} = \sqrt{\frac{\mathrm{T}_1}{\mathrm{T}_2}} = \sqrt{\frac{346}{800}} = \frac{1}{2} $
This means:
$ \mathrm{v}_2 = 2 \mathrm{v}_1 $
So, option (B) is the only correct answer.

Answer

The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules.

Answer

The temperature of gas is $-73^{\circ} \mathrm{C}$. When the gas is heated to $527^{\circ} \mathrm{C}$, the r.m.s. speed of the molecules is doubled.

Answer

The temperature of gas is $-100^{\circ} \mathrm{C}$. When the gas is heated to $+627^{\circ} \mathrm{C}$, the r.m.s. speed of the molecules is four times.

Answer

The product of pressure and volume of an ideal gas will be equal to half the translational kinetic energy.

Thermodynamics & KTG

Select the correct statement.

Choose the correct answer: