A sample P of an ideal gas contains 1 mol at an absolute temperature T - AQA - A-Level Physics - Question 9 - 2020 - Paper 2

The molecular kinetic energy (KE) of an ideal gas is given by the equation:

$KE = \frac{3}{2} nRT$

where:

• ( n ) is the number of moles,
• ( R ) is the universal gas constant,
• ( T ) is the absolute temperature.

For sample P:

• Number of moles, ( n_P = 1 )
• Temperature, ( T_P = T )
• Thus, the total kinetic energy for P is:

$KE_P = \frac{3}{2} (1)RT = \frac{3}{2} RT = E$

For sample Q:

• Number of moles, ( n_Q = \frac{2}{3} )
• Temperature, ( T_Q = 2T )
• Thus, the total kinetic energy for Q is:

$KE_Q = \frac{3}{2} \left( \frac{2}{3} \right) R(2T) = \frac{3}{2} \cdot \frac{2}{3} \cdot 2RT = 2RT$

Now, using the relation found earlier that ( E = \frac{3}{2} RT ), we replace ( RT ) in the expression for ( KE_Q ):

$KE_Q = 2RT = 2 \cdot \frac{2E}{3} = \frac{4E}{3}$

Thus, the total molecular kinetic energy of sample Q is ( \frac{4}{3}E ).