Verified Solution • 2025–26 Session moderate

QJEEM17Phee6a: JEE Main 2017 Physics - Class 11 JEEM17Phee6a Solution: The temperature of an open room of volume 30 m 3 increases from …

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 :

Multiple Choice Options

  • A - 1.61 ×\times 1023
  • B 1.38 ×\times 1023
  • C 2.5 ×\times 1025
  • D - 2.5 ×\times 1025

Step-by-Step Verified Solution

Given: Initial temperature Ti = 17 + 273 = 290 K

Final temperature Tf = 27 + 273 = 300 K

Atmospheric pressure, P0 = 1 × 105 Pa

Volume of room, V0 = 30 m3

Difference in number of molecules, Nf – Ni = ?

We know PV = nRT = NNA{N \over {{N_A}}} RT

The number of molecules

N = PVNART{{PV{N_A}} \over {RT}}

∴\therefore Nf – Ni = P0V0NAR(1Tf−1Ti){{{P_0}{V_0}{N_A}} \over R}\left( {{1 \over {{T_f}}} - {1 \over {{T_i}}}} \right)

= 1×105×30×6.023×10238.314(1300−1290){{1 \times {{10}^5} \times 30 \times 6.023 \times {{10}^{23}}} \over {8.314}}\left( {{1 \over {300}} - {1 \over {290}}} \right)

= – 2.5 ×\times 1025

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