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 · 8 Apr · Shift 1 · Q22
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 · 8 Apr · Shift 1 · Q22

Thermodynamics question

2019 · 8 Apr · Shift 1 · Q22

JEE MainChemistryThermodynamicsMCQ+4 / −1
For silver, Cp(J K–1 mol–1) = 23 +0.01 T. If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of ΔH\Delta HΔH will be close to :
  1. A
    62 KJ
  2. B
    16 KJ
  3. C
    13 KJ
  4. D
    21 KJ
View written solutionFree

Correct answer: A

  1. Use the definition of enthalpy change at constant pressure

For a substance heated at constant pressure,

ΔH=n∫T1T2Cp(T) dT\Delta H = n\int_{T_1}^{T_2} C_p(T)\, dTΔH=n∫T1​T2​​Cp​(T)dT

Given:

  • Cp=23+0.01T  J K−1mol−1C_p = 23 + 0.01T\; \text{J K}^{-1}\text{mol}^{-1}Cp​=23+0.01TJ K−1mol−1
  • n=3n=3n=3 moles
  • T1=300 KT_1=300\,\text{K}T1​=300K
  • T2=1000 KT_2=1000\,\text{K}T2​=1000K

So,

ΔH=3∫3001000(23+0.01T) dT\Delta H = 3\int_{300}^{1000} (23+0.01T)\, dTΔH=3∫3001000​(23+0.01T)dT
  1. Integrate
∫(23+0.01T) dT=23T+0.01⋅T22=23T+0.005T2\int (23+0.01T)\, dT = 23T + 0.01\cdot \frac{T^2}{2} = 23T + 0.005T^2∫(23+0.01T)dT=23T+0.01⋅2T2​=23T+0.005T2

Therefore,

ΔH=3[23T+0.005T2]3001000\Delta H = 3\left[23T+0.005T^2\right]_{300}^{1000}ΔH=3[23T+0.005T2]3001000​
  1. Substitute the limits

At T=1000T=1000T=1000:

23(1000)+0.005(1000)2=23000+5000=2800023(1000)+0.005(1000)^2 = 23000 + 5000 = 2800023(1000)+0.005(1000)2=23000+5000=28000

At T=300T=300T=300:

23(300)+0.005(300)2=6900+450=735023(300)+0.005(300)^2 = 6900 + 450 = 735023(300)+0.005(300)2=6900+450=7350

Difference:

28000−7350=20650  J mol−128000 - 7350 = 20650\; \text{J mol}^{-1}28000−7350=20650J mol−1

For 3 moles:

ΔH=3×20650=61950  J\Delta H = 3\times 20650 = 61950\; \text{J}ΔH=3×20650=61950J ΔH=61.95  kJ\Delta H = 61.95\; \text{kJ}ΔH=61.95kJ
  1. Closest option
ΔH≈62  kJ\Delta H \approx 62\; \text{kJ}ΔH≈62kJ

So the correct option is A.

PreviousNext

More from Thermodynamics

  • 5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV = 28 JK–1mol–1, calculate Δ U and Δ pV for this process. (R = 8.0 JK–1 mol–1]2019 · MCQ
  • Among the following, the set of parameters that represents path function, is : (A) q + w (B) q (C) w (D) H–TS2019 · MCQ
  • During compression of a spring the work done is 10kJ and 2kJ escaped to the surroundings as heat. The change in internal energy, Δ U(inkJ) is :2019 · MCQ
  • 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 :2019 · MCQ
  • The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is : (Specific heat of water liquid and water vapour are 4.2 kJ K − 1 kg − 1 and 2.0 kJ K − 1 kg − 1; heat of liquid fusion and…2019 · MCQ
  • A process will be spontaneous at all temperatures if :2019 · MCQ
  • The difference between Δ H and Δ U (Δ H – Δ U), when the combustion of one mole of heptane(l) is carried out at a temperature T, is equal to :2019 · MCQ
  • A process has Δ H = 200 J mol–1 and Δ S = 40 JK–1 mol–1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous :2019 · MCQ