AP EAMCET · PHYSICS · Kinetic Theory of Gases
In a container of volume \(16.62 \mathrm{~m}^3\) at \(0{ }^{\circ} \mathrm{C}\) temperature, 2 moles of oxygen, 5 moles of nitrogen and 3 moles of hydrogen are present, then the pressure in the container is
\(\left(\right.\) Universal gas constant \(\left.=8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)\)
- A 1570 Pa
- B 1270 Pa
- C 1365 Pa
- D 2270 Pa
Answer & Solution
Correct Answer
(C) 1365 Pa
Step-by-step Solution
Detailed explanation
\(n_{total} = 2 + 5 + 3 = 10 \mathrm{~mol}\) \(P = \frac{n_{total}RT}{V} = \frac{10 \mathrm{~mol} \times 8.31 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} \times 273.15 \mathrm{~K}}{16.62 \mathrm{~m}^3} = 1365.80 \mathrm{~Pa}\)
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