Not so hard if you remember Grahams Law of diffusion through a small hole. For these gases the relative loss would be sqrt(32/28) or about 1.07. So the answer is A.Answer 2
The rate of effusion (or the speed that the molecules will leave the spacecraft) is directly related to the molecule's molar mass, so immediately you can rule out C, D, and E. Now to do the math.
g/mol 02= 32
Graham's law states that the effusion rate of a molecule is equivalent to the inverse square root of it's mass. So...
The squrare root of 32/28= 1.069
The answer is A.Answer 3
C) The use of Graham's law is not relevant here. The gasses do not "diffuse" or "effuse" they are forced out into the vacuum of space at a terrific speed - the outside pressure being nearly zero atmospheres.
Since the partial pressures of the gases are equal ( .5 atm each) their relative amounts (moles) are equal so it is a 50-50 mixture of N2 and O2.
The key to this question is the fact that the pressure on the other side of the opening (no matter how small) is near zero.