THERMAL O6

Physics Solution Q6

Solution

The temperature of an ideal gas is proportional to the average kinetic energy of its molecules.

$$ T \propto \langle v^2 \rangle $$

Let’s calculate the sum proportional to Total Energy initially ($\sum N_i v_i^2$) using percentage as $N$.

Removed Speed Classes

Initial State ($N=100$):
$\sum N v^2 \propto 10(1^2) + 20(4^2) + 40(6^2) + 20(8^2) + 10(10^2)$ (using units of 100 m/s)
Sum = $10 + 320 + 1440 + 1280 + 1000 = 4050$
Average $\langle v^2 \rangle_{initial} = \frac{4050}{100} = 40.5$

Given $T_{initial} = 324 \text{ K}$. So proportionality constant $k = \frac{324}{40.5} = 8$.

Final State (Removing 800 m/s):
We remove 20 molecules of speed 8 (contribution $1280$).
New Sum = $4050 – 1280 = 2770$
New Total N = $100 – 20 = 80$
Average $\langle v^2 \rangle_{final} = \frac{2770}{80} = 34.625$

New Temperature $T_{final} = k \times 34.625 = 8 \times 34.625 = 277 \text{ K}$.

Change in Temperature = $277 – 324 = -47 \text{ K}$.

Correct Option: (b) Decrease by 47 $^\circ$C