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AIIMS 2015
Exam
AIIMS
Year
2015
Subject
Physics
Chapter
Kinetic Theory of Gases

AIIMS 2015 · Physics · Kinetic Theory of Gases

1 mole of H_2 gas is contained in a box of volume V=1.00 ~m^3 at...

1 mole of H2\mathrm{H}_2 gas is contained in a box of volume V=1.00 m3V=1.00 \mathrm{~m}^3 at T=300 KT=300 \mathrm{~K}. The gas is heated to a temperature of T=3000 KT=3000 \mathrm{~K} and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

Options

  1. Asame as the pressure initially
  2. B2 times the pressure initially
  3. C10 times the pressure initially
  4. D20 times the pressure initially

Correct answer

D. 20 times the pressure initially

Step-by-step solution

According to gas equation P V=n R T or P2V2T2=P1V1T1\quad \frac{P_2 V_2}{T_2}=\frac{P_1 V_1}{T_1} \quad or P2P1=V1V2⋅T2T1\quad \frac{P_2}{P_1}=\frac{V_1}{V_2} \cdot \frac{T_2}{T_1} Here, T2=3000 K,T1=300 KT_2=3000 \mathrm{~K}, T_1=300 \mathrm{~K} Since H2\mathrm{H}_2 splits into hydrogen atoms, therefore volume become half i.e., V2=12V1V_2=\frac{1}{2} V_1∴P2P1=V112V1×3000300 or P2P1=20\begin{aligned}& \therefore \quad \frac{P_2}{P_1}=\frac{V_1}{\frac{1}{2} V_1} \times \frac{3000}{300}\\ & \text{ or } \frac{P_2}{P_1}=20 \end{aligned}$
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