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Question

# 4 g of 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 50∘C more that the first one, 0.8 g of the gas had to be removed to get the original pressure. T is equal to:

A
510 K
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B
200 K
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C
100 K
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D
73 K
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Solution

## The correct option is B 200 K4 g of argon gas is to be maintained at the same temperature and pressure i.e., when temperature is raised, excess argon gas is removed. So using Ideal gas equation, PV = nRT PV is conatant in both the conditions So, n1RT1=n2RT2 n1=440 = 0.1 mole n2=3.240 = 0.08 moles initial temperature is T1 K and Final temperature is T1+50 K equating all values 0.1×T1=0.08×(T1+50) solving we get T1 = 200 K

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