- T1 is lower than T2
- The rms speed at T1 is higher than that at T2.
- The peak of each graph shows the most probable speed at the corresponding temperature.
- None of the above.
This is College Physics Answers with Shaun Dychko. This graph shows the distribution of molecular speeds in a gas at two different temperatures. Now temperature two is going to be greater than temperature one because we can see that that there are higher speeds in this distribution for temperature two. And the peak velocity of the most probable velocity is higher for temperature two than it is for temperature one. And increased velocity is evidence of increased kinetic energy of the particles and that is basically where temperature is measuring. So, we know that <i>T1</i> is lower than <i>T2</i>. So <i>T1</i> is less than <i>T2</i> in other words. That's good. By the way, that's not our answer. We're just going to put a check mark beside that, so that is not false. The <i>rms</i> speed at <i>T1</i> is higher than that at <i>T2</i>. That seems unlikely because at peak speed at <i>T1</i> is lower than the peak speed at <i>T2</i>. Now, this point is higher than this point is but don't let that confuse you because the height, the vertical access in other words is just a measuring probability. And so, at temperature one it is more probable to find a molecule at the peak velocity than it is to find a molecule at temperature two at the peak velocity. That part is true, but nevertheless the peak velocity at temperature two is greater than the peak velocity at temperature one. And since the peak velocity at temperature two is greater than the peak velocity of temperature one, it also means that the <i>rms</i> velocity which is somewhere a little bit more than the peak velocity, it is <i>Vrms</i> for temperature two and here is <i>Vrms</i> for temperature one approximately. So that's the comparison there. And so the <i>rms</i> speed for temperature two is greater. So that means this is false. So that's going to be the answer to our question. But let's just confirm by showing that c is correct. The peak of each graph shows the most probable speed at the corresponding temperature. Yes, definitely. So there we go. All done.