back to directory
NEET PHYSICSWAVEMedium

Question

4.0 gm4.0 \text{ gm} of gas occupies 22.4 litres22.4 \text{ litres} at NTP. The specific heat capacity of the gas at a constant volume is 5.0 J K1mol15.0 \text{ J K}^{-1}\text{mol}^{-1}. If the speed of sound in the gas at NTP is 952 ms1952 \text{ ms}^{-1}, then the molar heat capacity at constant pressure will be: (R=8.31 J K1mol1R=8.31 \text{ J K}^{-1}\text{mol}^{-1})

A

8.0 J K1mol18.0 \text{ J K}^{-1}\text{mol}^{-1}

B

7.5 J K1mol17.5 \text{ J K}^{-1}\text{mol}^{-1}

C

7.0 J K1mol17.0 \text{ J K}^{-1}\text{mol}^{-1}

D

8.5 J K1mol18.5 \text{ J K}^{-1}\text{mol}^{-1}

Step-by-Step Solution

At NTP (Normal Temperature and Pressure), 22.4 L22.4 \text{ L} of an ideal gas corresponds to 1 mole1 \text{ mole}. Therefore, the molar mass of the gas is M=4.0 g/mol=4.0×103 kg/molM = 4.0 \text{ g/mol} = 4.0 \times 10^{-3} \text{ kg/mol}. The speed of sound in a gas is given by the formula: v=γRTMv = \sqrt{\frac{\gamma RT}{M}} Squaring both sides and solving for γ\gamma: γ=v2MRT\gamma = \frac{v^2 M}{RT} Given v=952 ms1v = 952 \text{ ms}^{-1}, R=8.31 J K1mol1R = 8.31 \text{ J K}^{-1}\text{mol}^{-1}, and T=273 KT = 273 \text{ K}: γ=(952)2×4.0×1038.31×273=906304×0.0042268.631.6\gamma = \frac{(952)^2 \times 4.0 \times 10^{-3}}{8.31 \times 273} = \frac{906304 \times 0.004}{2268.63} \approx 1.6 Alternatively, using density ρ=MassVolume=4.0×103 kg22.4×103 m3\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{4.0 \times 10^{-3} \text{ kg}}{22.4 \times 10^{-3} \text{ m}^3} and pressure P=1.013×105 PaP = 1.013 \times 10^5 \text{ Pa}: γ=v2ρP=(952)2×(4.022.4)1.013×1051.6\gamma = \frac{v^2 \rho}{P} = \frac{(952)^2 \times \left(\frac{4.0}{22.4}\right)}{1.013 \times 10^5} \approx 1.6 The ratio of molar heat capacities is γ=CpCv\gamma = \frac{C_p}{C_v}. Given the molar heat capacity at constant volume Cv=5.0 J K1mol1C_v = 5.0 \text{ J K}^{-1}\text{mol}^{-1}: Cp=γCv=1.6×5.0=8.0 J K1mol1C_p = \gamma C_v = 1.6 \times 5.0 = 8.0 \text{ J K}^{-1}\text{mol}^{-1}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE. 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.

PHYSICSWAVEoccupieslitresspecificcapacityconstant

More WAVE Questions

View all

If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:

A.$1:2$
B.$1:1$
C.$\sqrt{2}:1$
D.$1:\sqrt{2}$
EasySolve

When a string is divided into three segments of lengths $l_1, l_2$ and $l_3$, the fundamental frequencies of these three segments are $\nu_1, \nu_2$ and $\nu_3$ respectively. The original fundamental frequency ($\nu$) of the string is:

A.$\sqrt{\nu}=\sqrt{\nu_1}+\sqrt{\nu_2}+\sqrt{\nu_3}$
B.$\nu=\nu_1+\nu_2+\nu_3$
C.$\frac{1}{\nu}=\frac{1}{\nu_1}+\frac{1}{\nu_2}+\frac{1}{\nu_3}$
D.$\frac{1}{\sqrt{\nu}}=\frac{1}{\sqrt{\nu_1}}+\frac{1}{\sqrt{\nu_2}}+\frac{1}{\sqrt{\nu_3}}$
MediumSolve

Two periodic waves of intensities $I_1$ and $I_2$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is:

A.$I_1+I_2$
B.$(\sqrt{I_1}+\sqrt{I_2})^2$
C.$(\sqrt{I_1}-\sqrt{I_2})^2$
D.$2(I_1+I_2)$
EasySolve

The two nearest harmonics of a tube closed at one end and open at the other end are $220 \text{ Hz}$ and $260 \text{ Hz}$. What is the fundamental frequency of the system?

A.$10 \text{ Hz}$
B.$20 \text{ Hz}$
C.$30 \text{ Hz}$
D.$40 \text{ Hz}$
MediumSolve

A transverse wave propagating along the $x$-axis is represented by: $y(x,t) = 8.0\sin(0.5\pi x - 4\pi t - \frac{\pi}{4})$, where $x$ is in meters and $t$ in seconds. The speed of the wave is:

A.$4\pi \text{ m/s}$
B.$0.5 \text{ m/s}$
C.$\frac{\pi}{4} \text{ m/s}$
D.$8 \text{ m/s}$
EasySolve

The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is $20 \text{ cm}$, the length of the open organ pipe is:

A.$13.2 \text{ cm}$
B.$8 \text{ cm}$
C.$12.5 \text{ cm}$
D.$16 \text{ cm}$
MediumSolve

The number of possible natural oscillations of the air column in a pipe closed at one end of a length of $85 \text{ cm}$ whose frequencies lie below $1250 \text{ Hz}$ is: (velocity of sound $340 \text{ ms}^{-1}$)

A.4
B.5
C.7
D.6
MediumSolve

A source of unknown frequency gives $4 \text{ beats/s}$ when sounded with a source of known frequency $250 \text{ Hz}$. The second harmonic of the source of unknown frequency gives five beats per second when sounded with a source of frequency $513 \text{ Hz}$. The unknown frequency is

A.$254 \text{ Hz}$
B.$246 \text{ Hz}$
C.$240 \text{ Hz}$
D.$260 \text{ Hz}$
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 →