Question

A spacecraft is filled with 0.500 atm of N2 and 0.500 atm of O2. Suppose a micrometeor strikes this spacecraft and puts a very small hole in its side. Under these circumstances,

A. O2 is lost from the craft 6.9% faster than N2 is lost.

B. O2 is lost from the craft 14% faster than N2 is lost.

C. N2 is lost from the craft 6.9% faster than O2 is lost.

D. N2 is lost from the craft 14% faster than O2 is lost.

E. N2 and O2 are lost from the craft at the same rate.

Answer 1

C. N₂ is lost from the craft 6.9% faster than O₂ is lost.

Step-by-step explanation:

The rate of effusion of a gas (r) is inversely proportional to the square root of its molar mass (M), as expressed by the Graham's law.

`\frac{rN_(2)}{r_{O_(2)}} =\sqrt{(M(O_(2)))/(M(N_(2))) } =\sqrt{(32.00g/mol)/(28.00g/mol) } =1.069\\rN_(2)=1.069rO_(2)`

The rate of effusion of N₂ is 1.069 times the rate of effusion of O₂, that is, N₂ effuses 6.9% faster than O₂.

Answer 2

N2 is lost from the craft 6.9% faster than O2 is lost.

Step-by-step explanation:

Step 1: Data given

0.500 atm of N2

0.500 atm of 02

Molar mass of 02 = 2*16 g/mole = 32 g/mole

Molar mass of N2 = 2* 14 g/mole = 28g/mole

Step 2: Calculate the rate of loss

r1N2 / r2O2 = √(M2(02)/M1(N2)) = √(32/28) = 1.069

1.069-1)*100= 6.9%

This means N2 is lost from the craft 6.9% faster than O2 is lost.