__About me__

I am a student at the University of Cambridge where I study Natural Sciences, a course which allows me to maintain my interest in multiple areas of science and maths. I take modules in Chemistry, Physics, Earth Science and Maths and am able and willing to help in any of those areas.

I have found from being taught in supervisions at university that the best thing you can take from the experience is an appreciation for the subject so I aim to try to convey the joy I find in the sciences and hope that some of my enthusiasm rubs off on you.

__Supervisions__

I have had some tutoring experience before so I know that an understanding of the subject is far more valuable than simply knowing the textbook answers so this will be my focus. However, I will defer to you for the content you wish to cover and will be happy to help with exam questions if that is what you are looking for.

By the end of a supervision, I will ensure that you are confident enough in your knowledge of the topic that you can explain it to me via whichever method you find most useful (be it diagrams, words or maths).

__What next__

I you have any questions or think I can help you, feel free to contact me or book a session, both through the website. It will help if I know your exam board and what I can help with.

I look forward to helping you.

__About me__

I am a student at the University of Cambridge where I study Natural Sciences, a course which allows me to maintain my interest in multiple areas of science and maths. I take modules in Chemistry, Physics, Earth Science and Maths and am able and willing to help in any of those areas.

I have found from being taught in supervisions at university that the best thing you can take from the experience is an appreciation for the subject so I aim to try to convey the joy I find in the sciences and hope that some of my enthusiasm rubs off on you.

__Supervisions__

I have had some tutoring experience before so I know that an understanding of the subject is far more valuable than simply knowing the textbook answers so this will be my focus. However, I will defer to you for the content you wish to cover and will be happy to help with exam questions if that is what you are looking for.

By the end of a supervision, I will ensure that you are confident enough in your knowledge of the topic that you can explain it to me via whichever method you find most useful (be it diagrams, words or maths).

__What next__

I you have any questions or think I can help you, feel free to contact me or book a session, both through the website. It will help if I know your exam board and what I can help with.

I look forward to helping you.

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

No DBS Check

First, write the line in vector form __r__=(1 1 1) + t(1 2 -1). Consider a point P on the line such that the line connecting P and A is perpendicular to L. The vector P->A is (2 3 4) - (1 1 1) - t(1 2 -1) = (1 2 3) - t(1 2 -1).

To make P->A perpendicular to L, the dot product of P->A and the direction vector of L must be zero. This means (1 2 -1) . (1-t 2-2t 3+t)=0 so 1-t+4-4t-3-t=0 or 2-6t=0 so t=1/3. This means that P is (4/3 7/3 -2/3) and P->A is (2/3 4/3 10/3). Therefore the shortest distance between A and L is sqrt(4/9+16/9+100/9)=2sqrt(10/3).

First, write the line in vector form __r__=(1 1 1) + t(1 2 -1). Consider a point P on the line such that the line connecting P and A is perpendicular to L. The vector P->A is (2 3 4) - (1 1 1) - t(1 2 -1) = (1 2 3) - t(1 2 -1).

To make P->A perpendicular to L, the dot product of P->A and the direction vector of L must be zero. This means (1 2 -1) . (1-t 2-2t 3+t)=0 so 1-t+4-4t-3-t=0 or 2-6t=0 so t=1/3. This means that P is (4/3 7/3 -2/3) and P->A is (2/3 4/3 10/3). Therefore the shortest distance between A and L is sqrt(4/9+16/9+100/9)=2sqrt(10/3).

To start with, some assumptions must be made to simplify the problem. Firstly the N molecules are assumed to have no size, be identical and not interact with each other. The second assumption is that the molecules move randomly and collide elastically with the container which is a cube of side length L.

When a particle collides with the container in the x direction, it rebound in the opposite direction at the same speed meaning the change in momentum is 2mv_{x}. The momentum change of n particles is therefore 2nmv_{x}.

The force exerted on the wall F=Impulse/Time so next we need to find the number of particles colliding with the wall per second. To do this, consider that all particles less than v metres from the wall will collide before one second. This means that any particle in the volume v_{x}L^{2} will collide. However, only half the particles will be travelling towards the wall so a factor of 1/2 is needed. The number of particles in this volume is v_{x}L^{2}/2*N/L^{3}=v_{x}N/2L. This means F=2mv_{x}+v_{x}N/2L=mNv_{x}^{2}/L. As all directions are equivalent, v^{2}=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}=3v_{x}^{2} so F=mNv^{2}/3L.

The pressure is therefore P=F/L^{2}=mNv^{2}/3L^{3}=mNv^{2}/3V. This means that PV=Nmv^{2}/3. Also, PV=Nk_{B}T so Nk_{B}T=Nmv^{2}/3 so k_{B}T=mv^{2}/3. The average speed can now be found in terms of temperature and mass only: v=sqrt(3k_{B}T/m).

To start with, some assumptions must be made to simplify the problem. Firstly the N molecules are assumed to have no size, be identical and not interact with each other. The second assumption is that the molecules move randomly and collide elastically with the container which is a cube of side length L.

When a particle collides with the container in the x direction, it rebound in the opposite direction at the same speed meaning the change in momentum is 2mv_{x}. The momentum change of n particles is therefore 2nmv_{x}.

The force exerted on the wall F=Impulse/Time so next we need to find the number of particles colliding with the wall per second. To do this, consider that all particles less than v metres from the wall will collide before one second. This means that any particle in the volume v_{x}L^{2} will collide. However, only half the particles will be travelling towards the wall so a factor of 1/2 is needed. The number of particles in this volume is v_{x}L^{2}/2*N/L^{3}=v_{x}N/2L. This means F=2mv_{x}+v_{x}N/2L=mNv_{x}^{2}/L. As all directions are equivalent, v^{2}=v_{x}^{2}+v_{y}^{2}+v_{z}^{2}=3v_{x}^{2} so F=mNv^{2}/3L.

The pressure is therefore P=F/L^{2}=mNv^{2}/3L^{3}=mNv^{2}/3V. This means that PV=Nmv^{2}/3. Also, PV=Nk_{B}T so Nk_{B}T=Nmv^{2}/3 so k_{B}T=mv^{2}/3. The average speed can now be found in terms of temperature and mass only: v=sqrt(3k_{B}T/m).

At first glance, this may look like an inverse trig integral but as the top contains an x term, we must use a different method.

First, rewrite the top in two parts, one that is a multiple of the derivative of the bottom and one that is just a number. In this case the derivative of the bottom is 2x+2 so rewrite as (2x+2)/2+3.

Now split the fraction in two: (2x+2)/2(x^{2}+2x+2)+3/(x^{2}+2x+2). The first part is now a log integral integrating to 1/2 ln(x^{2}+2x+2).

To integrate the second we must complete the square on the bottom. x^{2}+2x+2=(x+1)^{2}+1. Now using the substitution u=x+1, dx=du, we can see that the second part is an inverse tan integral integrating to arctan(x+1).

The overall integral is therefore 1/2 ln(x^{2}+2x+2) + arctan(x+1).

At first glance, this may look like an inverse trig integral but as the top contains an x term, we must use a different method.

First, rewrite the top in two parts, one that is a multiple of the derivative of the bottom and one that is just a number. In this case the derivative of the bottom is 2x+2 so rewrite as (2x+2)/2+3.

Now split the fraction in two: (2x+2)/2(x^{2}+2x+2)+3/(x^{2}+2x+2). The first part is now a log integral integrating to 1/2 ln(x^{2}+2x+2).

To integrate the second we must complete the square on the bottom. x^{2}+2x+2=(x+1)^{2}+1. Now using the substitution u=x+1, dx=du, we can see that the second part is an inverse tan integral integrating to arctan(x+1).

The overall integral is therefore 1/2 ln(x^{2}+2x+2) + arctan(x+1).