Take free online mock test for National Institute of Open Schooling (NIOS) Class XII Physics (312) chapter 'Kinetic Theory of Gases'. 20 MCQs, 20 minutes, detailed solutions & score analysis.
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Mastering Kinetic Theory of Gases in Class 12 provides critical subject intuition, analytical skills, and examination readiness.
Examinations test fundamental principles, definitions, cause-and-effect reasoning, and real-world applications under timed conditions.
Grasping core concepts and practicing with timed chapter-wise CBT tests guarantees top performance in board exams.
| Term / Formula | Type | Definition / Meaning | Explanation / Notes | Example / Usage |
|---|---|---|---|---|
| Core Concept | Principle | Primary concept governing Kinetic Theory of Gases. | मुख्य सिद्धांत | "Understand and apply the principle accurately." |
| Analytical Deduction | Skill | Evaluating evidence to draw correct conclusions. | तार्किक विश्लेषण | "Examine question clues systematically." |
Here is a sneak peek of the questions included in this test set. Test your preparation by viewing the answer and detailed explanation before starting the timed exam.
The root-mean-square (rms) speed of oxygen gas molecules ($\text{O}_2$, molar mass $M = 32.0\text{ g/mol} = 0.032\text{ kg/mol}$) at an absolute temperature of $T = 300\text{ K}$ is: (Take $R = 8.314\text{ J/(mol}\cdot\text{K)}$)
$T = 300\text{ K}$ के पूर्ण तापमान पर ऑक्सीजन गैस अणुओं ($\text{O}_2$, दाढ़ द्रव्यमान $M = 32.0\text{ g/mol} = 0.032\text{ kg/mol}$) की मूल-माध्य-वर्ग (आरएमएस) गति है: ($R = 8.314\text{ J/(mol}\cdot\text{K)}$ लें)
$484\text{ m/s}$
$384\text{ m/s}$
$515\text{ m/s}$
$420\text{ m/s}$
Step-by-step solution:
1. The root-mean-square speed of ideal gas molecules is given by:
$$v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}$$
2. Substitute $R = 8.314\text{ J/(mol}\cdot\text{K)}$, $T = 300\text{ K}$, and $M = 0.032\text{ kg/mol}$:
$$v_{\text{rms}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}} = \sqrt{\frac{7482.6}{0.032}} = \sqrt{233{,}831.25} \approx 483.6\text{ m/s} \approx 484\text{ m/s}$$
चरण-दर-चरण समाधान:
1. आदर्श गैस अणुओं की मूल-माध्य-वर्ग गति इस प्रकार दी गई है:
$$v_{\text{rms}} = \sqrt{\frac{3 R T}{M}}$$
2. विकल्प $R = 8.314\text{ J/(mol}\cdot\text{K)}$, $T = 300\text{ K}$, और $M = 0.032\text{ kg/mol}$:
$$v_{\text{rms}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}} = \sqrt{\frac{7482.6}{0.032}} = \sqrt{233{,}831.25} \approx 483.6\text{ m/s} \approx 484\text{ m/s}$$
The average translational kinetic energy of a single gas molecule at a room temperature of $27.0^\circ\text{C}$ ($300\text{ K}$) is: (Take Boltzmann's constant $k_B = 1.38 \times 10^{-23}\text{ J/K}$)
कमरे के तापमान $27.0^\circ\text{C}$ ($300\text{ K}$) पर एकल गैस अणु की औसत स्थानान्तरणीय गतिज ऊर्जा है: (बोल्ट्जमैन स्थिरांक लें $k_B = 1.38 \times 10^{-23}\text{ J/K}$)
$4.14 \times 10^{-21}\text{ J}$
$6.21 \times 10^{-21}\text{ J}$
$2.07 \times 10^{-21}\text{ J}$
$8.28 \times 10^{-21}\text{ J}$
Step-by-step solution:
1. By the equipartition theorem, the average translational kinetic energy per molecule in three dimensions is:
$$\bar{\epsilon}_{\text{trans}} = \frac{3}{2} k_B T$$
2. Substitute $k_B = 1.38 \times 10^{-23}\text{ J/K}$ and $T = 300\text{ K}$:
$$\bar{\epsilon}_{\text{trans}} = \frac{3}{2} \times (1.38 \times 10^{-23}\text{ J/K}) \times 300\text{ K} = 6.21 \times 10^{-21}\text{ J}$$
Note: This quantity depends solely on absolute temperature and is identical for all ideal gases regardless of molecular mass.
चरण-दर-चरण समाधान:
1. समविभाजन प्रमेय के अनुसार, तीन आयामों में प्रति अणु औसत अनुवादात्मक गतिज ऊर्जा है:
$$\bar{\epsilon}_{\text{trans}} = \frac{3}{2} k_B T$$
2. विकल्प $k_B = 1.38 \times 10^{-23}\text{ J/K}$ और $T = 300\text{ K}$:
$$\bar{\epsilon}_{\text{trans}} = \frac{3}{2} \times (1.38 \times 10^{-23}\text{ J/K}) \times 300\text{ K} = 6.21 \times 10^{-21}\text{ J}$$
नोट: यह मात्रा पूरी तरह से पूर्ण तापमान पर निर्भर करती है और आणविक द्रव्यमान की परवाह किए बिना सभी आदर्श गैसों के लिए समान है।
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