# Celsius to Kelvin conversion problem (Kinetic theory)

Forco

## Homework Statement

Find the temperature T that allows the rms speed of a gas to be equal to another gas with T=47°C.
The molecular mass of the first gas is 64, and the molecular mass of the second gas is 32.

## Homework Equations

$$v_{rms}= \sqrt{\frac{3RT}{M}}$$

## The Attempt at a Solution

The problem is actually very easy. It's actually really simple to conclude that
$$T_1=2T_2$$. However, my problem arises when actually replacing the given temperature.
If I take the second temperature to equal 47°C, then the first temperature is equal to 94°C. And converting that to kelvin gives 367.15 K.
However, if instead I use directly the temperature in K (47+273.15), then my answer becomes 640.3 K.
Which one is right? I assume the second one because in order for the equation to make sense, T needs to be expressed in K. I'd like to be sure, however.

Mentor
The problem is actually very easy. It's actually really simple to conclude that
$$T_1=2T_2$$.
Well, that depends which gas you label 1, and which you label 2.

However, my problem arises when actually replacing the given temperature.
If I take the second temperature to equal 47°C, then the first temperature is equal to 94°C. And converting that to kelvin gives 367.15 K.
However, if instead I use directly the temperature in K (47+273.15), then my answer becomes 640.3 K.
Which one is right? I assume the second one because in order for the equation to make sense, T needs to be expressed in K. I'd like to be sure, however.