1.

0.4 gm of He in a bulb at a temperature of 'T' K had a pressure of 'P' atm. When the bulb was immersed in hotter bath at a temperature 50K more than the first one, 0.08 gm of gas had to be removed to restore the original pressure. Then value of 'T' is:

Answer»

100K
200K
300K
500K

Solution :Since P and V is CONSTANT
`n_(i)T_(i)=n_(F)T_(f)`
`(0.4)/(4)xx(T)=((0.4-0.08))/(4)xx(T+50)`
`5T=4(T+50)`
T=200K


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