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Thermodynamics question

2019 · 9 Jan · Shift 1 · Q11
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Thermodynamics question

2019 · 9 Jan · Shift 1 · Q11

JEE MainChemistryThermodynamicsMCQ+4 / −1
Consider the reversible isothermal expansion of an ideal gas in a closed system at two different temperatures T1 and T2 (T1 < T2). The correct graphical depiction of the dependence of work done (w) on the final volume (V) is :
  1. A
    JEE Main 2019 (Online) 9th January Morning Slot Chemistry - Thermodynamics Question 161 English Option 1
  2. B
    JEE Main 2019 (Online) 9th January Morning Slot Chemistry - Thermodynamics Question 161 English Option 2
  3. C
    JEE Main 2019 (Online) 9th January Morning Slot Chemistry - Thermodynamics Question 161 English Option 3
  4. D
    JEE Main 2019 (Online) 9th January Morning Slot Chemistry - Thermodynamics Question 161 English Option 4
View written solutionFree

Correct answer: B

  1. Work done in reversible isothermal expansion of an ideal gas

For a reversible isothermal expansion of an ideal gas from initial volume ViV_iVi​ to final volume VVV at temperature TTT,

w=−∫ViVP dVw = -\int_{V_i}^{V} P\,dVw=−∫Vi​V​PdV

Using the ideal gas equation for a reversible path,

P=nRTVP = \frac{nRT}{V}P=VnRT​

So,

w=−∫ViVnRTV dV=−nRTln⁡(VVi)w = -\int_{V_i}^{V} \frac{nRT}{V}\,dV = -nRT \ln\left(\frac{V}{V_i}\right)w=−∫Vi​V​VnRT​dV=−nRTln(Vi​V​)
  1. Dependence of www on final volume VVV

Thus,

w(V)=−nRTln⁡(VVi)w(V) = -nRT \ln\left(\frac{V}{V_i}\right)w(V)=−nRTln(Vi​V​)

For expansion, V>ViV > V_iV>Vi​, so ln⁡(V/Vi)>0\ln(V/V_i) > 0ln(V/Vi​)>0, hence

w<0w < 0w<0

So the graph lies below the VVV-axis.

  1. Shape of the graph

Differentiate with respect to VVV:

dwdV=−nRTV<0\frac{dw}{dV} = -\frac{nRT}{V} < 0dVdw​=−VnRT​<0

Hence www decreases with increasing VVV.

Second derivative:

d2wdV2=nRTV2>0\frac{d^2w}{dV^2} = \frac{nRT}{V^2} > 0dV2d2w​=V2nRT​>0

So the graph is concave upward.

Also, at V=ViV = V_iV=Vi​,

w=−nRTln⁡1=0w = -nRT\ln 1 = 0w=−nRTln1=0

Thus the curve starts from w=0w=0w=0 and goes downward into negative values, flattening gradually.

  1. Comparison for two temperatures T1T_1T1​ and T2T_2T2​ with T1<T2T_1 < T_2T1​<T2​

For the same final volume VVV,

w1=−nRT1ln⁡(VVi),w2=−nRT2ln⁡(VVi)w_1 = -nRT_1 \ln\left(\frac{V}{V_i}\right), \qquad w_2 = -nRT_2 \ln\left(\frac{V}{V_i}\right)w1​=−nRT1​ln(Vi​V​),w2​=−nRT2​ln(Vi​V​)

Since T2>T1T_2 > T_1T2​>T1​,

∣w2∣>∣w1∣|w_2| > |w_1|∣w2​∣>∣w1​∣

That means the higher-temperature curve lies below the lower-temperature curve (more negative work) for the same VVV.

  1. Conclusion about the correct graph

So the correct graph must show:

  • both curves starting at w=0w=0w=0 when V=ViV=V_iV=Vi​,
  • both curves going into negative values as VVV increases,
  • concave upward shape,
  • the T2T_2T2​ curve below the T1T_1T1​ curve.

This corresponds to Option B.

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