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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2017
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  • MCQ

The temperature of an open room of volume \(30\ m^3\) increases from \(17^o C\) to \(27vC\) due to sunshine. The atmospheric pressure in the room remains \(1 \times 10^5\ Pa\). Ifni and nr are the number of molecules in the room before and after heating, then \(n_f-n_i\) will be :

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Answer & Solution

Correct Answer

(D) \(-2.5 \times 10^{25}\)

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Detailed explanation

\(\text { Given: Temperature } T_{i}=17+273=290 \mathrm{K}\)

Temperature \(T_{f}=27+273=300 K\)

Atmospheric pressure, \(P_{0}=1 \times 10^{5} P a\)

\( Volume\, of\, room,\) \(V_{0}=30 \mathrm{m}^{3}\)

\( Difference \,in\, number\, of\, molecules,\) \(N_{f}-N_{i}=?\)

The number of molecules

\(\Rightarrow \quad N=\frac{P V}{R T}\left(\mathrm{N}_{0}\right)\)

\(\begin{aligned} \therefore \quad N_{f}-N_{i} =\frac{P_{0} V_{0}}{R}\left(\frac{1}{T_{f}}-\frac{1}{T_{i}}\right) \mathrm{N}_{0} \\ =\frac{1 \times 10^{5} \times 30}{8.314} \times 6.023 \times 10^{23}\left(\frac{1}{300}-\frac{1}{290}\right) \\ =-2.5 \times 10^{25} \end{aligned}\)

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