back to directory
NEET PHYSICSKinetic TheoryEasy

Question

The kinetic energy of one gram molecule of a gas at standard temperature and pressure is: (R = 8.31 J/mol-K)

A

0.56 × 10^4 J

B

1.3 × 10^2 J

C

2.7 × 10^2 J

D

3.4 × 10^3 J

Step-by-Step Solution

The kinetic energy (E) of 'one gram molecule' (which equals 1 mole) of an ideal gas is given by the formula E=32RTE = \frac{3}{2}RT. Given: R=8.31 J mol1 K1R = 8.31 \text{ J mol}^{-1} \text{ K}^{-1} . Standard Temperature (STP) T=273 KT = 273 \text{ K} (\approx 0C0^\circ\text{C}). Substituting the values: E=32×8.31×273E = \frac{3}{2} \times 8.31 \times 273. E=1.5×2268.63=3402.9 JE = 1.5 \times 2268.63 = 3402.9 \text{ J}. In scientific notation, this is approximately 3.4×103 J3.4 \times 10^3 \text{ J}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Kinetic Theory. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSKinetic Theorykineticenergymoleculestandardtemperature

More Kinetic Theory Questions

View all

The mean free path l for a gas molecule depends upon the diameter, d of the molecule as:

A.l ∝ 1/d²
B.l ∝ d
C.l ∝ d²
D.l ∝ 1/d
EasySolve

The value $\gamma = \frac{C_P}{C_V}$ for hydrogen, helium, and another ideal diatomic gas $X$ (whose molecules are not rigid but have an additional vibrational mode), are respectively equal to:

A.$\frac{7}{5}, \frac{5}{3}, \frac{9}{7}$
B.$\frac{5}{3}, \frac{7}{5}, \frac{9}{7}$
C.$\frac{5}{3}, \frac{7}{5}, \frac{7}{5}$
D.$\frac{7}{5}, \frac{5}{3}, \frac{7}{5}$
MediumSolve

An ideal gas at $0^\circ\text{C}$ and atmospheric pressure $P$ has volume $V$. The percentage increase in its temperature needed to expand it to $3V$ at constant pressure is:

A.100%
B.200%
C.300%
D.50%
EasySolve

The root-mean-square speed of hydrogen molecules at 300 K is 1930 m/s. Then the root mean square speed of oxygen molecules at 900 K will be:

A.1930√3 m/s
B.836 m/s
C.643 m/s
D.1930/√3 m/s
MediumSolve

The molar specific heats of an ideal gas at constant pressure and volume are denoted by $C_P$ and $C_V$ respectively. If $\gamma = C_P/C_V$ and $R$ is the universal gas constant, then $C_V$ is equal to:

A.(1+\gamma)/(1-\gamma)
B.R/(\gamma-1)
C.(\gamma-1)/R
D.\gamma R
EasySolve

In the given V-T diagram, what is the relation between pressure $P_1$ and $P_2$?

A.$P_2 > P_1$
B.$P_2 < P_1$
C.cannot be predicted
D.$P_2 = P_1$
EasySolve

One mole of an ideal monatomic gas undergoes a process described by the equation $PV^3 = \text{constant}$. The heat capacity of the gas during this process is:

A.3/2 R
B.5/2 R
C.2 R
D.R
MediumSolve

The volume occupied by the molecules contained in $4.5 \text{ kg}$ water at STP, if the molecular forces vanish away, is:

A.$5.6 \text{ m}^3$
B.$5.6 \times 10^6 \text{ m}^3$
C.$5.6 \times 10^3 \text{ m}^3$
D.$5.6 \times 10^{-3} \text{ m}^3$
MediumSolve

This neet physics practice question is part of the TopperSquare free question bank. TopperSquare offers 15,000+ chapter-wise NEET MCQs across Physics, Chemistry, and Biology with detailed step-by-step explanations, full mock tests, NEET PYQs (2010–2024), and an AI-powered performance analytics dashboard. browse all neet practice questions → · practice physics sets →