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NEET 2011
Exam
NEET
Year
2011
Subject
Physics
Chapter
Thermodynamics

NEET 2011 · Physics · Thermodynamics

A mass of diatomic gas (γ=1.4) at a pressure of 2 ~atm is...

A mass of diatomic gas (γ=1.4)(\gamma=1.4) at a pressure of 2 atm2 \mathrm{~atm} is compressed adiabatically so that its temperature rise from 27∘C27^{\circ} \mathrm{C} to 927∘C927^{\circ} \mathrm{C}. The pressure of the gas is final state is

Options

  1. A28 atm28 \mathrm{~atm}
  2. B68.7 atm68.7 \mathrm{~atm}
  3. C256 atm256 \mathrm{~atm}
  4. D8 atm8 \mathrm{~atm}

Correct answer

C. 256 atm256 \mathrm{~atm}

Step-by-step solution

T1=273+27=300KT2=273+927=1200K\begin{aligned} & T_1=273+27=300 \mathrm{ K} \\ & T_2=273+927=1200 \mathrm{ K} \end{aligned} Gas equation for adiabatic process pVγ= constant p(Tp)γ= constant (∵pV=RT)∴p2p1=(T2T1)γγ−1p2=p1(T2T1)γγ−1p2=2(1200300)1.41.4−1p2=256atm\begin{aligned} & p V^\gamma=\text{ constant } \\ & p\left(\frac{T}{p}\right)^\gamma=\text{ constant } \quad(\because p V=R T) \\ & \therefore \quad \frac{p_2}{p_1}=\left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}} \\ & p_2=p_1\left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}} \\ & p_2=2\left(\frac{1200}{300}\right)^{\frac{1.4}{1.4-1}} \\ & p_2=256 \mathrm{ atm} \end{aligned}
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