Tag: principal and molar specific heats of gases

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Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

If water at ${ 0 }^{ \circ  }C.$kept in a container with an open top , is placed in a large evacuated chamber- 

  1. All the water will sported

  2. All the water will French .

  3. Part of the water will vaporize will be formed and reached . equilibrium at the triple point.

  4. ice , water and vapour will be formed and reach equilibrium at the triple points .
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

Equal volumes of monoatomic and diatomic gases of same initial temperature and pressure are mixed. The ratio of the specific heats of the mixture ($C _p/C _v$) will be

  1. $1.53$

  2. $1.52$

  3. $1.5$

  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$C _p$ for a gas is given as $(1+\displaystyle\dfrac{f}{2})R$ and $C _v$ is given as $\displaystyle\dfrac{f}{2}R$. Here f is the degree of freedom. 


For monoatomic gas it is 3 and for a diatomic gas it is 5. 

Thus, we get $C _p$ for the mixture as $\displaystyle\dfrac{5}{2}R+\dfrac{7}{2}R=6R$ and $C _v$ is given as $\displaystyle\dfrac{3}{2}R+\dfrac{5}{2}R=4R$.

Thus, the ratio $\displaystyle\dfrac{C _p}{C _v}$ is given as $\displaystyle\dfrac{3}{2}=1.5$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The quantity of heat (in J) required to raise the temperature of $1.0\, kg$ of ethanol from $293.45\, K$ to the boiling point and then change the liquid to vapor at that temperature is closest to 
[Given : Boiling point of ethanol $351.45\, K$
              Specific heat capacity of liquid ethanol $2.44\, J\, g^{-1}\, K^{-1}$
               Latent heat of vaporization of ethanol $855\, J \, g^{-1}$]

  1. $1.42\, \times 10^2$

  2. $9.97\, \times 10^2$

  3. $1.42\, \times 10^5$

  4. $9.97\, \times 10^5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Heat to raise temperature: Q1 = m * c * delta T = 1000g * 2.44 J/gK * (351.45 - 293.45)K = 1000 * 2.44 * 58 = 141,520 J. Heat to vaporize: Q2 = m * L = 1000g * 855 J/g = 855,000 J. Total Q = 141,520 + 855,000 = 996,520 J, which is approximately 9.97 * 10^5 J.