p= nRT/v

P and V are related by the equation  where all the other quantities in
the equation are constant. What is the gradient of a graph in which P
is plotted against 1/V?

Asking you to find the gradient is just another way of asking you to
determine the slope of the line that results from plotting (p) vs.
(1/v).

It is important to note that p is not necessarily equal to 1/v, but it
will always be proportional to it, since the other variables in the
equation are constants.

Let's assume nRT=10.  Therefore, if P is 5, v must be 2, and 1/v equals 1/2.

Now, still assuming nRT=10 (since it is always constant), if P = 10, v
must be 1, and 1/v = 1.

One last example... if P = 2, v = 5, and 1/v = 1/5.

So our coordinates are (assuming P is on the x-axis, and 1/v is on the
y-axis)...  (2, 1/5) and (5, 1/2) and (10, 1)

It should be clear from these data that the slope is 1/10.  Using the
first two data points, as the y-value rises by 3/10, the x-value rises
by 3.  The slope is equal to the change in the y-value divided by the
change in the x-value, so the slope = (3/10) / (3) = 1/10 = 0.1

If you find the slope using the other two combinations of the above
data points, the result is also 0.1.

One important point... saying P against 1/V does not make clear which
value is supposed to be plotted on which axis.  The 0.1 answer is true
if you were supposed to plot P on the x-axis and 1/v on the y-axis. 
If it was the other way around, the slope would be the reciprocal of
0.1, which is 10.

For what you are asking the gradient will be nRT, this is because you
are plotting against 1/v, which makes the graph have a constant
gradient.

The function

p= nRT/v  with n, R, T being constants

is not a line.  It is a hyperbola.  So the slope will vary at
different places along the curve.  The slope can be determined by
taking the derivative of the function.

d(nRT/v)/dv = -nRT/v^2

At v=1/2 the slope is -4nRT
At v=1 the slope is -nRT
At v=2 the slope is -nRT/4

I am assuming that the vertical axis is p and the horizonal axis is v.
 The two axes are the asmyptotes to the curve.