 # If we triple the average kinetic energy of the gas atoms, what is the new temperature in ∘c? Assume that the ideal gas is at 40C.The aim of this question is to understand the relationship between temperature and kinetic energy of ideal gas molecules.

The formula for the average kinetic energy of an ideal gas is:

$E \ = \ \dfrac{ 3 }{ 2 } k_b T$

Where,

$E \ = \ \text{ average kinetic energy }, \ k_b \ = \ \text{ Boltzmann constant }, \ T \ = \ \text{ temperature }$

Notice that temperature and kinetic energy are directly proportional.

The average kinetic energy of an ideal gas can be calculated using the following formula:

$E \ = \ \dfrac{ 3 }{ 2 } k_b T$

Rearranging:

$\dfrac{ E }{ \dfrac{ 3 }{ 2 } k_b } \ = \ T$

$\Rightarrow T \ = \ \dfrac{ 2 E }{ 3 k_b } \ … \ … \ … \ (1)$

Given:

$T \ = \ 40^{ \circ } \ = \ 40 \ + \ 273.15 \ = \ 313.15 \ K$

Substituting in above equation (1):

$313.15 \ K \ = \ \dfrac{ 2 E }{ 3 k_b } \ … \ … \ … \ (2)$

Now if we triple the kinetic energy:

$E \ \rightarrow \ 3 E$

Then equation (1) for new temperature value $T’$ becomes:

$T’ \ = \ \dfrac{ 2 ( \ 3 E \ ) }{ 3 k_b }$

Rearranging:

$T’ \ = \ 3 \bigg ( \dfrac{ 2 E }{ 3 k_b } \bigg )$

Substituting value of $\dfrac{ 2 E }{ 3 k_b }$ from equation (2):

$T’ \ = \ 3 \bigg ( \ 313.15 \ K \ \bigg )$

$\Rightarrow T’ \ = \ 939.45 \ K$

$\Rightarrow T’ \ = \ 939.45 \ – \ 273.15 \ ^{ \circ } C$

$\Rightarrow T’ \ = \ 666.30 ^{ \circ } C$

## Numerical Result

$T’ \ = \ 666.30 ^{ \circ } C$

## Example

If we double the average kinetic energy of the gas atoms, what is the new temperature in ∘c? Assume that the ideal gas is at $\boldsymbol{ 20^{ \circ } C }$.

Recall equation (1):

$T \ = \ \dfrac{ 2 E }{ 3 k_b }$

Given:

$T \ = \ 20^{ \circ } \ = \ 20 \ + \ 273.15 \ = \ 293.15 \ K$

Substituting in above equation (1):

$293.15 \ K \ = \ \dfrac{ 2 E }{ 3 k_b } \ … \ … \ … \ (3)$

Now if we double the kinetic energy:

$E \ \rightarrow \ 2 E$

Then equation (1) for new temperature value $T^{ ” }$ becomes:

$T^{ ” } \ = \ \dfrac{ 2 ( \ 2 E \ ) }{ 3 k_b }$

Rearranging:

$T^{ ” } \ = \ 2 \bigg ( \dfrac{ 2 E }{ 3 k_b } \bigg )$

Substituting value of $\dfrac{ 2 E }{ 3 k_b }$ from equation (3):

$T’ \ = \ 2 \bigg ( \ 293.15 \ K \ \bigg )$

$\Rightarrow T’ \ = \ 586.30 \ K \ = \ 586.30 \ – \ 273.15 \ ^{ \circ } C \ = \ 313.15 ^{ \circ } C$