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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Physics
Chapter
Thermodynamics

JEE Advanced 2011 · Paper 1 · Physics · Thermodynamics

5.6 ~L of helium gas at STP is adiabatically compressed to 0.7 ~L ....

5.6 L5.6 \mathrm{~L} of helium gas at STP is adiabatically compressed to 0.7 L0.7 \mathrm{~L}. Taking the initial temperature to be T1T_1, the work done in the process is

Options

  1. A98RT1\frac{9}{8} R T_1
  2. B32RT1\frac{3}{2} R T_1
  3. C158RT1\frac{15}{8} R T_1
  4. D92RT1\frac{9}{2} R T_1

Correct answer

A. 98RT1\frac{9}{8} R T_1

Step-by-step solution

At STP, 22.4 L of any gas is 1 mole. \therefore \quad 5.6 L=\mathrm{ L}=5.622.4=14 moles =\frac{5.6}{22.4}=\frac{1}{4} \text{ moles }=n In adiabatic process, TVγ−1= constant ∴T2V2γ−1=T1V1γ−1 or T2=T1(V1V2)γ−1\begin{array}{rlrl} & T V^{\gamma-1} & =\text{ constant } \\ & \therefore & T_2 V_2^{\gamma-1} & =T_1 V_1^{\gamma-1} \\ & \text{ or } & T_2 & =T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1} \end{array}γ=CpCV=53 for monoatomic He gas. ∴T2=T1(5.60.7)53−1=4T1∵ Further in adiabatic process, Q=0∴W+ΔU=0 or W=−ΔU=−nCVΔT=−n(Rγ−1)(T2−T1)=−14(R53−1)(4T1−T1)=−98RT1\begin{aligned} & \gamma=\frac{C_p}{C_V}=\frac{5}{3} \text{ for monoatomic He gas. } \\ & \therefore \quad T_2=T_1\left(\frac{5.6}{0.7}\right)^{\frac{5}{3}-1}=4 T_1 \\ & \because \text{ Further in adiabatic process, } \\ & Q=0 \\ & \therefore \quad W+\Delta U=0 \\ & \text{ or } \quad W=-\Delta U \\ & =-n C_V \Delta T \\ & =-n\left(\frac{R}{\gamma-1}\right)\left(T_2-T_1\right) \\ & =-\frac{1}{4}\left(\frac{R}{\frac{5}{3}-1}\right)\left(4 T_1-T_1\right) \\ & =-\frac{9}{8} R T_1 \\ & \end{aligned} \thereforeCorrectoptionis(a).AnalysisofQuestion(i)Fromcalculationpointofviewquestionismoderatelytough.(ii)Volumeofgasisdecreasing.Therefore,workdonebythegasshouldbenegative.Correct option is (a). Analysis of Question (i) From calculation point of view question is moderately tough. (ii) Volume of gas is decreasing. Therefore, work done by the gas should be negative.
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