JEE MAIN - Physics (2017 (Offline) - No. 14)
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 :
- 1.61 $$ \times $$ 1023
1.38 $$ \times $$ 1023
2.5 $$ \times $$ 1025
- 2.5 $$ \times $$ 1025
Explanation
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 = $${N \over {{N_A}}}$$RT
The number of molecules
N = $${{PV{N_A}} \over {RT}}$$
$$ \therefore $$ Nf – Ni = $${{{P_0}{V_0}{N_A}} \over R}\left( {{1 \over {{T_f}}} - {1 \over {{T_i}}}} \right)$$
= $${{1 \times {{10}^5} \times 30 \times 6.023 \times {{10}^{23}}} \over {8.314}}\left( {{1 \over {300}} - {1 \over {290}}} \right)$$
= – 2.5 $$ \times $$ 1025
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 = $${N \over {{N_A}}}$$RT
The number of molecules
N = $${{PV{N_A}} \over {RT}}$$
$$ \therefore $$ Nf – Ni = $${{{P_0}{V_0}{N_A}} \over R}\left( {{1 \over {{T_f}}} - {1 \over {{T_i}}}} \right)$$
= $${{1 \times {{10}^5} \times 30 \times 6.023 \times {{10}^{23}}} \over {8.314}}\left( {{1 \over {300}} - {1 \over {290}}} \right)$$
= – 2.5 $$ \times $$ 1025
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