You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equipartition Theorem

Specific heat...

Question

Specific heat of a monoatomic gas at constant volume is $315 J k g^{- 1} K^{- 1}$ and at a constant pressure is $525 J k g^{- 1} K^{- 1}$ . Calculate the molar mass of the gas.

Open in App

Solution

$C_{P} = M \times 525$ and $C_{V} = M \times 315$
where, $M$ is the molecular mass.
$C_{P} - C_{V} = R (R = 8.314 K K^{- 1} m o l^{- 1})$
$M \times 525 - M \times 315 = 8.314$
$M (525 - 315) = 8.314$
$M = \frac{8.314}{210} = 0.0396 k g m o l^{- 1} = 39.6 g m o l^{- 1}$

Suggest Corrections

Similar questions

12.6 J m o l^{- 1} K^{- 1}

J m o l^{- 1} K^{- 1}

R = 8.314 J m o l^{- 1} K^{- 1})

1 k g

N T P

11.2 m^{3}

10046 J k g^{- 1} k^{- 1}

J k g^{- 1} k^{- 1}

10^{5} P a

1 k g

N . T . P

11.2 m^{3}

10046 J k g^{- 1} K^{- 1}

4150

J k g^{- 1} K^{- 1}

1.4

J k g^{- 1} K^{- 1}

0.075 k c a l k g^{- 1} K^{- 1}

Join BYJU'S Learning Program