back to directory
NEET PHYSICSWAVEMedium

Question

4.0 g4.0 \text{ g} of a gas occupies 22.4 L22.4 \text{ L} at NTP. The specific heat capacity of the gas at constant volume is 5.0 J K1mol15.0 \text{ J K}^{-1}\text{mol}^{-1}. If the speed of sound in this gas at NTP is 952 ms1952 \text{ ms}^{-1}, then the heat capacity at constant pressure is: (Take gas constant R=8.3 J K1mol1R = 8.3 \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

  1. Find Molar Mass (MM): At NTP (Normal Temperature and Pressure, T=273 K,P=1 atmT = 273 \text{ K}, P = 1 \text{ atm}), 1 mole1 \text{ mole} of an ideal gas occupies 22.4 L22.4 \text{ L}. Given that 4.0 g4.0 \text{ g} of the gas occupies 22.4 L22.4 \text{ L}, the molar mass of the gas is: M=4.0 g mol1=4.0×103 kg mol1M = 4.0 \text{ g mol}^{-1} = 4.0 \times 10^{-3} \text{ kg mol}^{-1}
  2. Calculate the Ratio of Specific Heats (γ\gamma): The speed of sound vv in a gas is given by Laplace's formula : v=γRTMv = \sqrt{\frac{\gamma R T}{M}} Squaring both sides and solving for γ\gamma: γ=v2MRT\gamma = \frac{v^2 M}{R T} Substitute the given values (v=952 ms1v = 952 \text{ ms}^{-1}, M=4.0×103 kg mol1M = 4.0 \times 10^{-3} \text{ kg mol}^{-1}, R=8.3 J K1mol1R = 8.3 \text{ J K}^{-1}\text{mol}^{-1}, T=273 KT = 273 \text{ K}): γ=(952)2×4.0×1038.3×273=906304×0.0042265.9=3625.2162265.91.6\gamma = \frac{(952)^2 \times 4.0 \times 10^{-3}}{8.3 \times 273} = \frac{906304 \times 0.004}{2265.9} = \frac{3625.216}{2265.9} \approx 1.6
  3. Calculate Heat Capacity at Constant Pressure (CpC_p): We know that the ratio of molar heat capacities is γ=CpCv\gamma = \frac{C_p}{C_v} . Given Cv=5.0 J K1mol1C_v = 5.0 \text{ J K}^{-1}\text{mol}^{-1}, we can find CpC_p: Cp=γ×Cv=1.6×5.0=8.0 J K1mol1C_p = \gamma \times 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.

PHYSICSWAVEoccupiesspecificcapacityconstantvolume

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 →