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0.40 g of helium in a bulb at a temperat...

0.40 g of helium in a bulb at a temperature of T K had a pressure of P atm. When the bulb was immersed in hotter bath at temperature 50 K more than the first one, 0.08 g of gas had to be removed to restore the original pressure. T is

A

100 K

B

200 K

C

300 K

D

500 K

Text Solution

Verified by Experts

The correct Answer is:
B
As P and V remains constant , `n_1 T_1 = n_2 T_2`
`n_ 1 = (0.4)/(4) = 0.1, T_1 = T K , n_(2) = (0.40-0.08)/(4) = 0.08`
`T_2 = (T+ 50)K`
On putting the values , `0.1 xx T = 0.08 xx (T + 50)`
`implies 0.1 T = 0.08 T + 4 implies 01 T - 0.08 T =4`
`implies 0.02 T = 4 implies T =200K`
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