Q40 / 180 · JIPMER 2015 · Chemistry

0.40 g of helium in a bulb at a temperature of $T \mathrm{~K}$ had a pressure of $p$ atm. When the bulb was immersed in water 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
    500 K
  • B
    400 K
  • C
    600 K
  • Correct answer
    D
    200 K

Solution

As pressure and volume remains constant, $\begin{aligned} n_1 T_1 & =n_2 T_2 \\ n_1 & =\frac{0.4}{4}=0.1, T_1=T \mathrm{~K} \\ n_2 & =\frac{0.40-0.08}{4}=0.08, T_2=(T+50) \mathrm{K}\end{aligned}$ On putting the values, $\begin{array}{ll} & 0.1 \times T=0.08 \times(T+50) \\ \text { or } & 0.1 \mathrm{~T}=0.08 T=4 \text { or } 0.1 \mathrm{~T}-0.08 \mathrm{~T}=4 \\ \text { or } & 0.02 T=4 \text { or } T=200 \mathrm{~K}\end{array}$

Hence, the correct answer is option: 'D'.

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