Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Thermodynamics question

2019 · 11 Jan · Shift 1 · Q16
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Thermodynamics
  5. /2019 · 11 Jan · Shift 1 · Q16

Thermodynamics question

2019 · 11 Jan · Shift 1 · Q16

JEE MainChemistryThermodynamicsMCQ+4 / −1
Two blocks of the same metal having same mass and at temperature T1 and T2, respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, Δ\DeltaΔ S, for this process is :
  1. A
    2Cp In [(T1+T2)12T1T2]\left[ {{{{{\left( {{T_1} + {T_2}} \right)}^{{1 \over 2}}}} \over {{T_1}{T_2}}}} \right][T1​T2​(T1​+T2​)21​​]
  2. B
    2Cp In [(T1+T2)2T1T2]\left[ {{{\left( {{T_1} + {T_2}} \right)} \over {2{T_1}{T_2}}}} \right][2T1​T2​(T1​+T2​)​]
  3. C
    Cp In [(T1+T2)24T1T2]\left[ {{{{{\left( {{T_1} + {T_2}} \right)}^2}} \over {4{T_1}{T_2}}}} \right][4T1​T2​(T1​+T2​)2​]
  4. D
    2Cp In [(T1+T2)4T1T2]\left[ {{{\left( {{T_1} + {T_2}} \right)} \over {4{T_1}{T_2}}}} \right][4T1​T2​(T1​+T2​)​]
View written solutionFree

Correct answer: C

  1. Final equilibrium temperature

Since the two blocks are of the same metal and have same mass, their heat capacities at constant pressure are equal. Let each block have heat capacity CpC_pCp​.

At thermal equilibrium, by energy balance:

Cp(Tf−T1)+Cp(Tf−T2)=0C_p(T_f-T_1)+C_p(T_f-T_2)=0Cp​(Tf​−T1​)+Cp​(Tf​−T2​)=0 2Tf=T1+T22T_f=T_1+T_22Tf​=T1​+T2​ Tf=T1+T22T_f=\frac{T_1+T_2}{2}Tf​=2T1​+T2​​
  1. Entropy change of each block

For a body with constant heat capacity CpC_pCp​, entropy change when temperature changes from TiT_iTi​ to TfT_fTf​ is

ΔS=Cpln⁡(TfTi)\Delta S=C_p\ln\left(\frac{T_f}{T_i}\right)ΔS=Cp​ln(Ti​Tf​​)

So for block 1:

ΔS1=Cpln⁡(TfT1)\Delta S_1=C_p\ln\left(\frac{T_f}{T_1}\right)ΔS1​=Cp​ln(T1​Tf​​)

For block 2:

ΔS2=Cpln⁡(TfT2)\Delta S_2=C_p\ln\left(\frac{T_f}{T_2}\right)ΔS2​=Cp​ln(T2​Tf​​)

Hence total entropy change:

ΔS=ΔS1+ΔS2\Delta S=\Delta S_1+\Delta S_2ΔS=ΔS1​+ΔS2​ ΔS=Cpln⁡(TfT1)+Cpln⁡(TfT2)\Delta S=C_p\ln\left(\frac{T_f}{T_1}\right)+C_p\ln\left(\frac{T_f}{T_2}\right)ΔS=Cp​ln(T1​Tf​​)+Cp​ln(T2​Tf​​) ΔS=Cpln⁡(Tf2T1T2)\Delta S=C_p\ln\left(\frac{T_f^2}{T_1T_2}\right)ΔS=Cp​ln(T1​T2​Tf2​​)
  1. Substitute Tf=T1+T22T_f=\dfrac{T_1+T_2}{2}Tf​=2T1​+T2​​
ΔS=Cpln⁡((T1+T22)2T1T2)\Delta S=C_p\ln\left(\frac{\left(\frac{T_1+T_2}{2}\right)^2}{T_1T_2}\right)ΔS=Cp​ln(T1​T2​(2T1​+T2​​)2​) ΔS=Cpln⁡((T1+T2)24T1T2)\Delta S=C_p\ln\left(\frac{(T_1+T_2)^2}{4T_1T_2}\right)ΔS=Cp​ln(4T1​T2​(T1​+T2​)2​)
  1. Match with options

This exactly matches:

Cpln⁡((T1+T2)24T1T2)\boxed{C_p\ln\left(\frac{(T_1+T_2)^2}{4T_1T_2}\right)}Cp​ln(4T1​T2​(T1​+T2​)2​)​

So the correct option is C.

  1. Check positivity

Since

(T1+T2)2≥4T1T2(T_1+T_2)^2 \ge 4T_1T_2(T1​+T2​)2≥4T1​T2​

we get

(T1+T2)24T1T2≥1\frac{(T_1+T_2)^2}{4T_1T_2} \ge 14T1​T2​(T1​+T2​)2​≥1

therefore ΔS≥0\Delta S\ge 0ΔS≥0, which is physically correct.

  1. Comparison with stored answer

Stored correct answer: C

Derived answer: C

They agree.

PreviousNext

More from Thermodynamics

  • The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by Δ rGo = A – BT Where A and B are non-zero constants. Which of the following is TRUE about this reaction?2019 · MCQ
  • The reaction, MgO(s) + C(s) → Mg(s) + CO(g), for which Δ rHo + 491.1 kJ mol–1 and Δ rSo = 198.0 JK–1 mol–1, is not feasible at 298 K. Temperature above which reaciton will be feasible is :2019 · MCQ
  • For the equilibrium, 2H2​O ⇌ H3​O+ + OH −, the value of Δ Go at 298 K is approximately :2019 · MCQ
  • An ideal gas is allowed to expand form 1 L to 10 L against a constant external pressure of I bar. The work done in kJ is :2019 · MCQ
  • Enthalpy of sublimation of iodine is 24 cal g–1 at 200 oC. If specific heat of I2​(s) and l2 (vap) are 0.055 and 0.031 cal g–1K –1 respectively, then enthalpy of sublimation of iodine at 250 oC in cal g–1 is :2019 · MCQ
  • For diatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities?2019 · MCQ
  • Given (i) C (graphite) + O2​(g) → CO2​(g); Δ rH Θ = x kJ mol − 1 (ii) C(graphite) + 21​ O2​(g) → CO(g); Δ rH Θ = y kJ mol − 1 (iii) CO(g) + 21​ O2​(g) →…2019 · MCQ
  • The combination of plots which does not represent isothermal expansion of an ideal gas is – Includes diagram2019 · MCQ