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JEE Advanced 2012Paper 2
Exam
JEE Advanced
Year
2012
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2012 · Paper 2 · Physics · Thermodynamics

Two moles of ideal helium gas are in a rubber balloon at 30^ C ....

Two moles of ideal helium gas are in a rubber balloon at 30∘C30^{\circ} \mathrm{C}. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35∘C35^{\circ} \mathrm{C}. The amount of heat required in raising the temperature is nearly (take R=8.31 J/mol.KR=8.31 \mathrm{~J} / \mathrm{mol} . \mathrm{K} )

Options

  1. A62 J62 \mathrm{~J}
  2. B104 J104 \mathrm{~J}
  3. C124 J124 \mathrm{~J}
  4. D208 J208 \mathrm{~J}

Correct answer

D. 208 J208 \mathrm{~J}

Step-by-step solution

The heat is supplied at constant pressure. i.e., the process is isobaric ∴Q=nCpΔt=2[52R]×Δt=2×52×8.31×5=208J(∵Cp=52R for mono-atomic gas )\begin{array}{l} \therefore Q=n C_{p} \Delta t \\ =2\left[\frac{5}{2} R\right] \times \Delta t=2 \times \frac{5}{2} \times 8.31 \times 5=208 \mathrm{ J} \\ \left(\because C_{p}=\frac{5}{2} R \text{ for mono-atomic gas }\right) \end{array}
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