Given condition: closed container, slowly heated

Since the gas is in a closed rigid container, we take:

• number of molecules $N$ = constant
• volume $V$ = constant

Hence, the number density $n = \frac{N}{V}$ remains constant.

Mean free path

For an ideal gas, the mean free path is $\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}$ where $d$ is molecular diameter and $n$ is number density.

Since $d$ is constant and $n$ is constant, we get $\lambda = \text{constant}$

So:

• (A) "mean free path decreases" = false
• (C) "mean free path remains unchanged" = true

Mean collision time

Mean collision time is $\tau = \frac{\lambda}{\bar v}$ where $\bar v$ is the average molecular speed.

For an ideal gas, $\bar v \propto \sqrt{T}$

As temperature increases, $\bar v$ increases. Since $\lambda$ remains constant, $\tau \propto \frac{1}{\sqrt{T}}$ so $\tau$ decreases.

Thus:

• (B) "mean collision time decreases" = true
• (D) "mean collision time remains unchanged" = false

Final evaluation of statements

The true statements are: $\boxed{(B) \text{ and } (C)}$

Matching with given options

Option C corresponds to $(B)$ and $(C)$.

Therefore, the correct answer is $\boxed{\text{C}}$