A closed container holds a sample of an ideal gas at 127^ C . To...

The initial temperature of the gas is $T_1 = 127^{\circ}\text{C} = 127 + 273 = 400\text{ K}$.

The root mean square speed of an ideal gas is given by $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$, which implies $v_{\text{rms}} \propto \sqrt{T}$.

Let the initial rms speed be $v_1$. Since the rms speed increases by $150\%$, the new rms speed $v_2$ is given by:

$v_2 = v_1 + 1.5v_1 = 2.5v_1$

Using the relation $v_{\text{rms}} \propto \sqrt{T}$, we have:

$\dfrac{v_2}{v_1} = \sqrt{\dfrac{T_2}{T_1}}$

Substituting the values:

$2.5 = \sqrt{\dfrac{T_2}{400}}$

Squaring both sides:

$6.25 = \dfrac{T_2}{400}$

$T_2 = 6.25 \times 400 = 2500\text{ K}$

Converting the temperature back to Celsius:

$T_2 = 2500 - 273 = 2227^{\circ}\text{C}$

Answer: $2227^{\circ}\text{C}$