Chemistry, asked by ayushichoubey7743, 8 months ago

Helium gas at 1 atm and SO2 at 2 atm pressure temperature being the same are released seperately at the same moment into 1 m long evacuated tubes of equal diameters If helium reaches the other end of the tube in t sec what distance SO2 would traverse in the same time interval in the other tube?

Answers

Answered by snehaaaaaaaaaa
5

Answer:

the given reactuon is :

2so3 reversible 2so2+o2

in this reaction a equation keeping temperature and volume constant helium gas is introduced so that total pressure is increase. the forward direction is accompanied by increase in number of Malls so if pressure is increased then as per Richard principle equilibrium shift in the direction of decreasing no number of moles soda dissociation of so3 is decrease.

Answered by Anonymous
1

Given:

  • The pressure of Helium gas = 1 atm.
  • The pressure of SO_2 = 2 atm.
  • Distance, (x) = 1 m = 100cm

To Find:

  • Distance covered by SO_2 to reach the other tube in the same interval of time.

Solution:

  • To find the distance covered by SO_2 to reach other tube with the same interval of time is given by, \frac{r_{He}}{r_{SO_2} }   = \frac{100/t}{x/t} = \frac{P_{He} }{P_{SO_2} }\sqrt{\frac{M_{SO_2} }{M_{He} } }
  • Where,
  • P_{He} is the pressure of He gas,
  • P_{SO_2} is the pressure of SO_2 gas,
  • M_{S0_2} is the molecular weight of SO_2,
  • M_{He} is the molecular weight of He gas.
  • x is the distance covered by SO_2 to other tube.
  • Substituting the given values, we get,
  • \frac{100}{x} = \frac{1}{2}\sqrt{\frac{64}{4} }
  • \frac{100}{x}= 0.5*\sqrt{16}
  • \frac{100}{x} = 0.5*4 = 2\\
  • 2x = 100
  • x = 100/2 = 50 cm.

The distance covered by SO_2 to other tube is 50 cm.

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