Question

5.6dm3 of gas requires 52.25 j of heat to raise it temperature by 10°C at constant volume gas can be?​

Answer 1

The value of constant gas will be γ=1.4

Explanation:

Consider the given values

Gas at STP=1 mol

5.6 $dm^{3}$ of the gas at S.T.P=$\frac{1}{22.4} *5.6$

=0.25mol

For $10^{0} C$ rise 0.25 mol of the gas at constant volume required heat =52.25J

For $1^{0}$ rise 1 mol of the gas at constant volume will required heat is

$=(52.25/10*0.25=20.9)$

$C_{v} =20.9 J K^{-1} mol^{-1}$

So

$C_{P}=C_{V} +R$

$=20.9JK^{-1} mol^{-1} +8.314JK^{-1} mol^{-1}$

$=29.21JK^{-1} mol^{-1}$

$r=\frac{C_{p} }{C_{v} }$

$=\frac{20.214}{20.9}$

$=1.4$

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