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Heat and Thermodynamics question

2017 · Shift 0 · Q57
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Heat and Thermodynamics question

2017 · Shift 0 · Q57

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The temperature of an open room of volume 30 m3 increases from 17oC to 27oC due to the sunshine. The atmospheric pressure in the room remains 1 ×\times× 105 Pa. If Ni and Nf are the number of molecules in the room before and after heating, then Nf – Ni will be :
  1. A
    - 1.61 ×\times× 1023
  2. B
    1.38 ×\times× 1023
  3. C
    2.5 ×\times× 1025
  4. D
    - 2.5 ×\times× 1025
View written solutionFree

Correct answer: D

  1. Use the ideal gas relation in molecular form

For an ideal gas, PV=NkTPV = NkTPV=NkT where:

  • PPP = pressure
  • VVV = volume
  • NNN = number of molecules
  • kkk = Boltzmann constant
  • TTT = absolute temperature

Since the room is open, pressure remains constant, and volume is also constant. Thus, N=PVkTN = \frac{PV}{kT}N=kTPV​ So the number of molecules is inversely proportional to temperature: N∝1TN \propto \frac{1}{T}N∝T1​

  1. Convert temperatures to Kelvin

Initial temperature: Ti=17∘C=290 KT_i = 17^\circ C = 290\,KTi​=17∘C=290K

Final temperature: Tf=27∘C=300 KT_f = 27^\circ C = 300\,KTf​=27∘C=300K

  1. Write expressions for NiN_iNi​ and NfN_fNf​

Ni=PVkTi,Nf=PVkTfN_i = \frac{PV}{kT_i}, \qquad N_f = \frac{PV}{kT_f}Ni​=kTi​PV​,Nf​=kTf​PV​

Hence, Nf−Ni=PVk(1Tf−1Ti)N_f - N_i = \frac{PV}{k}\left(\frac{1}{T_f} - \frac{1}{T_i}\right)Nf​−Ni​=kPV​(Tf​1​−Ti​1​)

  1. Substitute values

Given: P=1×105 Pa,V=30 m3,k=1.38×10−23 J/KP = 1\times 10^5\,Pa, \qquad V = 30\,m^3, \qquad k = 1.38\times 10^{-23}\,J/KP=1×105Pa,V=30m3,k=1.38×10−23J/K

First compute: PV=105×30=3×106PV = 10^5 \times 30 = 3\times 10^6PV=105×30=3×106

Now, 1Tf−1Ti=1300−1290\frac{1}{T_f} - \frac{1}{T_i} = \frac{1}{300} - \frac{1}{290}Tf​1​−Ti​1​=3001​−2901​ =290−300300×290=−1087000=−18700= \frac{290-300}{300\times 290} = \frac{-10}{87000} = -\frac{1}{8700}=300×290290−300​=87000−10​=−87001​

Therefore, Nf−Ni=3×1061.38×10−23⋅(−18700)N_f - N_i = \frac{3\times 10^6}{1.38\times 10^{-23}}\cdot \left(-\frac{1}{8700}\right)Nf​−Ni​=1.38×10−233×106​⋅(−87001​)

  1. Calculate

3×1068700≈344.83\frac{3\times 10^6}{8700} \approx 344.8387003×106​≈344.83

So, Nf−Ni≈−344.831.38×10−23N_f - N_i \approx -\frac{344.83}{1.38\times 10^{-23}}Nf​−Ni​≈−1.38×10−23344.83​ =−249.9×1023= -249.9\times 10^{23}=−249.9×1023 ≈−2.5×1025\approx -2.5\times 10^{25}≈−2.5×1025

  1. Final answer

Nf−Ni=−2.5×1025\boxed{N_f - N_i = -2.5\times 10^{25}}Nf​−Ni​=−2.5×1025​

So the correct option is D.

  1. Comparison with stored answer

Stored correct answer: D

Our derived answer also gives D, so they agree.

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