A vessel containing one mole of a monatomic ideal gas (molecular...

The kinetic energy of the gas due to the motion of the vessel is given by $K = \dfrac{1}{2} m v^2$.

The mass of the gas is $m = n \times M = 1 \times 20 \times 10^{-3} \text{ kg} = 0.02 \text{ kg}$.

Substituting the values, we get:

$K = \dfrac{1}{2} \times 0.02 \times (50)^2 = 25 \text{ J}$

When the vessel is abruptly brought to rest, this kinetic energy is entirely converted into the internal energy of the gas. Thus, the change in internal energy is $\Delta U = 25 \text{ J}$.

For a monatomic ideal gas, the change in internal energy is given by $\Delta U = n C_v \Delta T = n \left(\dfrac{3}{2} R\right) \Delta T$.

Equating the energies:

$25 = 1 \times \dfrac{3}{2} \times 8.31 \times \Delta T$

$25 = 12.465 \times \Delta T$

$\Delta T = \dfrac{25}{12.465} \approx 2.0 \text{ K}$