JEE Mains · Physics · STD 11 - Kinetic theory of gases
- JEE Mains 2017
- Medium
- 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 :
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(-2.5 \times 10^{25}\)
Step-by-step Solution
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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