Question

What is the temperature of 28 gm Co2 kept in a 20 liter container at 1.5 atm pressure​

Answer 1

Explanation:

Ideal gas equation

PV = nRT

Volume of container(V) remains same during leakage.

So initially

P1 = 10 atm

T1 = 57°C = 330 K

W1 = 28g

P1V = n1RT1 (1)

P2= 5 atm

T2 = 27°C = 300K

P2V = n2RT2 (2)

Dividing (1) by (2)

P1/P2 = n1T1/n2T2

10/5 = 330/300 × n1/n2

20/11 = n1/n2

But we know n1/n2=W1/W2

W1/W2 = 20/11

W2 = 28×11/20

W2 = 15.4g

Leaked mass = W1-W2 = 28-15.4 = 12.6g

Hence,12.6g N2 leaked out.

this method is used in this answer....

hope it will help you.......