pycrypto is slow encrypting and decrypting - python-2.7

Question

In practice, I select an executable. Size 20Mb.
I read the content using file.read(size=16).
If length of the returned byte string is less than 16 I fill the rest with \0 (NULL).
f = open("./installer.exe","rb")
obj = AES.new(b"0123456789012345",AES.MODE_CBC, b"\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0")
bs = b""
t = f.read(16)
while t != b"":
if len(t) < 16:
t = t + b"\0" * (16 - len(t)) # if < 16 bytes using padding
bs = bs + obj.encrypt(t)
else:
bs = bs + obj.encrypt(t)
t = f.read(16)
then, bs contents the byte string of ALL content encrypted with 0123456789012345
I realise the mechanism of reading file first, then I encrypt the content as seen in the above piece of code (using obj.encrypt()). Then I write a new file with the content encrypted. The I read the data of encrypted file and by a similar procedure decrypt the data using obj.decrypt in intervals of 16 bytes and then I write a new file with the decrypted data.
This takes approximately 3 minutes.
¿It's fast, slow, or expected?
According to what I saw, the module is written in C. ¿Maybe should I use Cython embedded to make it faster?
How PGP can supposedly decrypt higher amounts of data in real time, for example, in an encrypted virtual disk?
edit:
This take almost same:
obj = AES.new(b"0123456789012345",AES.MODE_CBC, b"\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0\0")
bs = b""
t = f.read(16)
while t != b"":
if len(t) < 16:
t = t + b"\0" * (16 - len(t))
bs = bs + t
else:
bs = bs + t
t = f.read(16)
bse = obj.encrypt(bs)

Answer 1

Ok. The problem was the size of the buffers encrypted. I decided to use 64000 bytes strings.
The procedure its simple. total size / string segments -> encrypt. And in the last segment, if the size of the segment is inferior to 64000 AND NOT multiple of 16, finds the nearest multiple and the remaining space is filled
bs = b""
dt = f.read()
dtl = len(dt)
dtr = ( dtl / 64000 ) + 1
for x in range(0, dtr):
if x == dtr-1:
i1 = 64000 * x
dst = dtl - i1
i = math.ceil(dst / 16.0) * 16
dst = i - dst
buf = dt[i1:] + (b"\0" * int(dst))
bs = bs + obj.encrypt(buf)
else:
i1, i2 = 64000 * x , 64000 * (x+1)
bs = bs + obj.encrypt(dt[i1:i2])
Now takes 10 seconds.
Thanks for all.