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'4'g argon (Atomic mass =40) in a bulb at a temperature of 'T'K has a pressure ‘P’ atm. When the bulb was placed in a hot bath at a temperature of 50°C more than the first one, 0.8g of gas had to be removed to get the original pressure. T is equal to:


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Solution

Ideal Gas Equation: PV=nRT

Where

  • P= pressure of the gas
  • V= volume of gas
  • n= mole of gas=GivenmassofthesubstanceAtomicmassofthesubstance
  • T= temperature of the gas
  • ‘R’ is a constant

As per the question,

P×V=440RT(Before placing the argon bulb in a hot bath)…………………equation(1)

P×V=(4-0.8)40R(T+50)(After placing the argon bulb in a hot bath)…..equation(2)

On equating equations (1)and (2).

440RT=(4-0.8)40R(T+50)

440T=3.240(T+50)

4T-3.2T=50×40=2000

0.8T=2000K

T=2500K

‘4'g argon (Atomic mass =40) in a bulb at a temperature of 'T'K has a pressure ‘P’ atm. When the bulb was placed in a hot bath at a temperature of 50°C more than the first one, 0.8g of gas had to be removed to get the original pressure. T is equal to 2500K.


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