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At a temperature T K, the pressure of 4.0 gm argon in a bulb is P. The bulb is put in a bath having temperature higher by 50 K then the first one. 0.8 gm of argon gas had to be removed to maintain original pressure. The temperature T is __

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510 K

200 K

100 K

73 K

answer is B.

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Detailed Solution

$T_{1} = TK T_{2} = 50 + T w_{1} = 4 g w_{2} = 4 - 0.8 = 3.2 g PV = constant from PV = \frac{w}{M} RT w_{1} T_{1} = w_{2} T_{2} \Rightarrow 4 \times T = 3.2 \times (50 + T) \Rightarrow T = 200 K$

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