Question

$\text{'}4\text{'}\mathrm{g}$ 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 $50°\mathrm{C}$ more than the first one, $0.8\mathrm{g}$ of gas had to be removed to get the original pressure. T is equal to:

Answer 1

Ideal Gas Equation: $\mathrm{PV}=\mathrm{nRT}$

Where

• P= pressure of the gas
• V= volume of gas
• n= mole of gas=$\frac{\mathrm{Given}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{substance}}{\mathrm{Atomic}\mathrm{mass}\mathrm{of}\mathrm{the}\mathrm{substance}}$
• T= temperature of the gas
• ‘R’ is a constant

As per the question,

$\mathrm{P}\times \mathrm{V}=\frac{4}{40}\mathrm{RT}$(Before placing the argon bulb in a hot bath)…………………equation(1)

$\mathrm{P}\times \mathrm{V}=\frac{(4-0.8)}{40}\mathrm{R}(\mathrm{T}+50)$(After placing the argon bulb in a hot bath)…..equation(2)

On equating equations (1)and (2).

$\frac{4}{40}\mathrm{RT}=\frac{(4-0.8)}{40}\mathrm{R}(\mathrm{T}+50)$

$\frac{4}{40}\mathrm{T}=\frac{3.2}{40}(\mathrm{T}+50)$

$4\mathrm{T}-3.2\mathrm{T}=50\times 40=2000$

$0.8\mathrm{T}=2000\mathrm{K}$

$\mathrm{T}=2500\mathrm{K}$

‘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.