Chemistry, asked by L2isaiv0asunaba, 1 year ago

A glass bulb of volume 400 cm3 is connected to another of volume 200 cm3 by means of a tube of negligible volume. The bulbs contain dry air and are both at a common temperature and pressure of 20.0C and 1.0 atm. The larger bulb is immersed in steam at 100 0C; the smaller, in melting ice at 0 0C. Find the final common pressure.

Answers

Answered by kvnmurty
43
V1 = 400 cm³     T = 293°K         P = 1 atm     n1 = 2 x
V2 = 200 cm³     T = 293°K          P = 1 atm     n2 = x
Total moles of the gas = 3 x   units

P V1 = n1 R T    => x = 200/(293 R)  units

After changing the temperatures:
    P' * 400 = n  R 373          and    P' 200 = (600 /293R  - n) * R 273

Taking their ratio:
  =>  2 (600/293R  - n) 273 = n 373
  =>  n = 1200* 273 /(293 * 919 * R)  units

P' = 1200 * 373 * 273 /(293 * 919 * 400)  = 1.134 atm

kvnmurty: click on thanks box/link above please
Answered by pravalikadaswgs
2

Answer:I dont know

Explanation:

Hehheheheheheh

Similar questions