If for hydrogen Cp-Cv=m and for nitrogen Cp-Cv= n where, C p and C v refer to specific heat per unit mass respectively at constant pressure and constant volume, the relation between m and n is (molecular weight of hydrogen =2 and molecular weight of nitrogen =28 )

Question

If for hydrogen Cp-Cv=m and for nitrogen Cp-Cv= n where, C p and C v refer to specific heat per unit mass respectively at constant pressure and constant volume, the relation between m and n is (molecular weight of hydrogen =2 and molecular weight of nitrogen =28 ) (A) n =14 m (B) n=7 m (C)  m=7 n  (D)  m=14 n

Tardigrade Admin · Accepted Answer

For hydrogen, C p - C v =( R / M H 2)=( R /2), but given that C n - C v = m m = R / 2 For nitrogen, Cp-Cv=n ∴ n = R / 28 ∴ ( m / n )=( R / 2/ R / 28)=14 Hence, m=14 n

$n = 14 m$

$n = 7 m$

$m = 7 n$

$m = 14 n$

For hydrogen,
$C_{p} - C_{v} = \frac{R}{M _{H 2}} = \frac{R}{2},$
but given that $C_{n} - C_{v} = m$
$m = R /2$
For nitrogen, $C_{p} - C_{v} = n$
$∴ n = R /28$
$∴ \frac{m}{n} = \frac{R /2}{R /28} = 14$
Hence, $m = 14 n$

Q. If for hydrogen Cp​−Cv​=m and for nitrogen Cp​−Cv​= n where, Cp​ and Cv​ refer to specific heat per unit mass respectively at constant pressure and constant volume, the relation between m and n is (molecular weight of hydrogen =2 and molecular weight of nitrogen =28 )

n=14m

n=7m

m=7n

m=14n

For hydrogen, Cp​−Cv​=MH2​R​=2R​, but given that Cn​−Cv​=m m=R/2 For nitrogen, Cp​−Cv​=n ∴n=R/28 ∴nm​=R/28R/2​=14 Hence, m=14n

$n = 14 m$

$n = 7 m$

$m = 7 n$

$m = 14 n$

For hydrogen,
$C_{p} - C_{v} = \frac{R}{M _{H 2}} = \frac{R}{2},$
but given that $C_{n} - C_{v} = m$
$m = R /2$
For nitrogen, $C_{p} - C_{v} = n$
$∴ n = R /28$
$∴ \frac{m}{n} = \frac{R /2}{R /28} = 14$
Hence, $m = 14 n$