The temperature of a gas is - 50 ° C . To what temperature the gas...

RMS speed is given by, $v = \sqrt{\dfrac{3RT}{M}}$.

Therefore,

$v ∝ \sqrt{T}$.

Given:

Initial temperature, ${T}_{1} = \left(- 50 + 273\right)K = 223K$

Let, initial speed be $v$, then final speed will be $v + 3v = 4v$.

Now,

$\dfrac{4v}{v} = \sqrt{\dfrac{{T}_{2}}{{T}_{1}}} ⇒ {T}_{2} = 16{T}_{1} = 3568K = \left(3568 - 273\right) ° C = 3295 ° C$