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Question: The density of a polyatomic gas is standard conditions is 0.795 \(kgm^{- 3}\). The specific heat of ...

The density of a polyatomic gas is standard conditions is 0.795 kgm−3kgm^{- 3}. The specific heat of the gas at constant volume is

A

930J⥂⥂−⥂kg−1K−1930J ⥂ ⥂ - ⥂ kg^{- 1}K^{- 1}

B

1400J⥂−⥂⥂kg−1K−11400J ⥂ - ⥂ ⥂ kg^{- 1}K^{- 1}

C

1120J⥂−⥂⥂kg−1K−11120J ⥂ - ⥂ ⥂ kg^{- 1}K^{- 1}

D

925J⥂−⥂kg−1K−1925J ⥂ - ⥂ kg^{- 1}K^{- 1}

Answer

1400J⥂−⥂⥂kg−1K−11400J ⥂ - ⥂ ⥂ kg^{- 1}K^{- 1}

Explanation

Solution

Ideal gas equation for m gram gas PV=mrTPV = mrT

[where r = Specific gas constant]

or P=mVrT=ρrTP = \frac{m}{V}rT = \rho rT

⇒ r=PρT=1.013×1050.795×273=466.7r = \frac{P}{\rho T} = \frac{1.013 \times 10^{5}}{0.795 \times 273} = 466.7

Specific heat at constant volume cv=rγ−1=466.743−1c_{v} = \frac{r}{\gamma - 1} = \frac{466.7}{\frac{4}{3} - 1} =1400Jkg.kelvin= 1400\frac{J}{kg.kelvin} [γ=43 for polyatomic gas]\left\lbrack \gamma = \frac{4}{3}\text{ for polyatomic gas} \right\rbrack