4.0 g of argon (at mass = 40) in a bulb at a temperature of TK had a pressure P atm. When the bulb was placed in hotter bath at a temperature `50^(@)` more than the first one, 0.8 g of gas had to be removed to get the original pressure. T is equal is

Options

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

Answer

B

Explanation

According to ideal gas eqatio
`PV = w (RT)/(M) " or " wT = (PV)/(R) .M`
or `w_(1)T_(1) = w_(2) T_(2)` when P and V are constant
Hence `4 xx T = 3.2 (T + 50)`
or `T = 200 K`