Here's an answer to the question as last modified.
First suppose that on the entire real line, not just on $[0,\infty),$ we assign to each interval $(a,b)$ a random variable $N_{(a,b)}$ for which
- $\Pr(N_{(a,b)} = n) = \dfrac{(\lambda(b-a))^n e^{-\lambda(b-a)}}{n!}$ for $n=0,1,2,3,\ldots,$ and
- For intervals $A,B,C,\ldots$ no two of which intersect each other, $N_A,N_B,N_C,\ldots$ are independent.
Then let $X_1$ be the time of the first arrival after time $0,$ let $X_2$ be the time from then until the next arrival after that, and so on. Then we have
$$
\Pr(X_n > x) = e^{-\lambda x} \text{ for } x\ge0.
$$
Now consider the time $Y$ from the last arrival before time $T>0$ until time $T.$ For $x\ge0$ we have
$$
\Pr(Y>x) = \Pr(\text{no arrivals during } [T-x,T]) = \frac{(\lambda x)^0 e^{-\lambda x}}{0!} = e^{-\lambda x}.
$$
I.e. it has the same distribution that any one of the inter-arrival times has.
And then if you want to truncate it at $x=T,$ as suggested in comments under the question, you can modify that accordingly.
If the day is of length $1$, and there are $n$ independent uniformly distributed start times then this is equivalent to breaking a stick into $n+1$ pieces and there is probably a duplicate question on this site.
The inter-arrival times are not independent but they are identically distributed (as is the time before the first arrival and the time after the last, so $n+1$ gaps overall). Jointly they have a uniform Dirichlet distribution.
Each gap actually has a $\text{Beta}(1,n)$ distribution so with CDF $1-(1-x)^n$ and density $n(1-x)^{n-1}$, with mean $\frac1{n+1}$ and variance $\frac{n}{(n+1)^2(n+2)}$. I have seen this called a "reverse power function distribution".
If the day is not length $1$ but $d$ then stretch these results so the CDF of the inter-arrival times is $1-\left(1-\frac xd\right)^{n}$ and density $\frac nd\left(1-\frac xd\right)^{n-1}$ and mean $\frac d{n+1}$ and variance $\frac{n d^2}{(n+1)^2(n+2)}$. For large $n$ this is close to an exponential distribution with rate $\frac{n+1}{d}$, as you might guess.
For example, here are the densities using R when $d=1$ and $n=100$, with the Beta distribution in red and exponential distribution in blue. They are very close to each other, and further right would be very close to $0$ (there is a slight distinction beyond $1$ as the Beta density would be exactly $0$ while the exponential density would be positive but less than $10^{−40}$). For larger n they would be even closer
n <- 100
curve(dbeta(x,1,n), from=0, to=6/n, col="red")
curve(dexp(x, n+1), from=0, to=6/n, add=TRUE, col="blue")
Best Answer
What do you mean by "inter arrival rate"?
If you mean that inter-arrival times are uniformly distributed, then the answer is no. The arrival process is not Poisson and the queue is a G/M/1.
The arrival process is only Poisson if the inter-arrival times follow an exponential distribution.