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:

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Solution

## Ideal Gas Equation: $\mathrm{PV}=\mathrm{nRT}$Where P= pressure of the gasV= volume of gasn= 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,$\mathrm{P}×\mathrm{V}=\frac{4}{40}\mathrm{RT}$(Before placing the argon bulb in a hot bath)…………………equation(1)$\mathrm{P}×\mathrm{V}=\frac{\left(4-0.8\right)}{40}\mathrm{R}\left(\mathrm{T}+50\right)$(After placing the argon bulb in a hot bath)…..equation(2)On equating equations (1)and (2).$\frac{4}{40}\mathrm{RT}=\frac{\left(4-0.8\right)}{40}\mathrm{R}\left(\mathrm{T}+50\right)$$\frac{4}{40}\mathrm{T}=\frac{3.2}{40}\left(\mathrm{T}+50\right)$$4\mathrm{T}-3.2\mathrm{T}=50×40=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.

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