For RSA, is there a specified size range for $p$ and $q$ when calculating $N$? Is there a specified size range for $p$ and $q$ when calculating $N$? For example, what size would $p$ and $q$ be when $N$ is 1024 bits?
There is no specific length for $p$ and $q$; however, one must mind the following:


Since $N = pq$, the size of $N$ will be more or less the sum of the sizes of $p$ and $q$ (± 1 bit).
Security of the key relies on the difficulty of integer factorization. The best known algorithms for that have a cost that depends on the size of $N$, but some other algorithms (in particular ECM) have a running cost that depends on the size of the smallest factor. If one of the factors $p$ and $q$ is too small, then it may make ECM practical, leading to a security weakening.
Private key operations with RSA can be expensive (especially on small, embedded hardware). A common optimisation is to do private key operations not modulo $N$, but modulo $p$ and $q$ (the Chinese remainder theorem tells us that it works). Since $p$ and $q$ are smaller than $N$, this is a net gain. The speed-up factor is highest when $p$ and $q$ are both about half the size of $N$.
Some widespread implementations of RSA require the two factors to be exactly half the size of the modulus. This is the case, for instance, for Microsoft's CryptoAPI: when importing a private key, the import format uses fixed-size slots for the factors, that are exactly half the size of the modulus.


Therefore, while the RSA standard does not mandate any specific sizes for the individual factors, the general practice is to make both factors $p$ and $q$ to be exactly half the size of the modulus $N$. So, for a 1024-bit modulus, the factors will be 512-bit integers.

(To be even more precise, it is customary to generate each factor with its two top bits set to 1. If you just take two 512-bit integers, their product may be a 1023-bit integer; by setting the two top bits of each factor to 1, this ensures that the product will be a 1024-bit integer.)