The ratio of specific heats for oxygen is 1.4. Density of oxygen at STP = text 1.44 text kg/m3. If atmospheric pressure is 1.013× 105N/m2, the specific heat at constant volume is:

Question

The ratio of specific heats for oxygen is 1.4. Density of oxygen at STP = text 1.44 text kg/m3. If atmospheric pressure is 1.013× 105N/m2, the specific heat at constant volume is: (A)  257.7 text J text kg-1K-1  (B)  644.3 text J text kg-1K-1  (C)  8.36 text J text kg-1K-1  (D)  901.2 text J text kg-1K-1

Tardigrade Admin · Accepted Answer

Given: (CP/CV)=1.4, T=273K ?(i) p=1.013× 105 N/m2 ρ =1.44 kg/m3 So, V=(1/1.44)m3 Also from Mayers formula, CP-CV=R=(PV/T) ?(ii) Putting the given values in Eq. (ii) and from Eq. (i) 1.4CV-CV=(1.013× 105× 1/1.44× 273) 0.4CV=257.7 CV=644.25 J/kg-K

Q. The ratio of specific heats for oxygen is 1.4. Density of oxygen at STP $= 1.44 k g / m^{3} .$ If atmospheric pressure is $1.013 \times 10^{5} N / m^{2},$ the specific heat at constant volume is:

$257.7 J k g^{- 1} K^{- 1}$

$644.3 J k g^{- 1} K^{- 1}$

$8.36 J k g^{- 1} K^{- 1}$

$901.2 J k g^{- 1} K^{- 1}$

Q. The ratio of specific heats for oxygen is 1.4. Density of oxygen at STP =1.44kg/m3. If atmospheric pressure is 1.013×105N/m2, the specific heat at constant volume is:

257.7Jkg−1K−1

644.3Jkg−1K−1

8.36Jkg−1K−1

901.2Jkg−1K−1

Given: CV​CP​​=1.4,T=273K ?(i) p=1.013×105N/m2 ρ=1.44kg/m3 So, V=1.441​m3 Also from Mayers formula, CP​−CV​=R=TPV​ ?(ii) Putting the given values in Eq. (ii) and from Eq. (i) 1.4CV​−CV​=1.44×2731.013×105×1​ 0.4CV​=257.7 CV​=644.25J/kg−K

Q. The ratio of specific heats for oxygen is 1.4. Density of oxygen at STP $= 1.44 k g / m^{3} .$ If atmospheric pressure is $1.013 \times 10^{5} N / m^{2},$ the specific heat at constant volume is:

$257.7 J k g^{- 1} K^{- 1}$

$644.3 J k g^{- 1} K^{- 1}$

$8.36 J k g^{- 1} K^{- 1}$

$901.2 J k g^{- 1} K^{- 1}$