Tag: heat and thermodynamics

Questions Related to heat and thermodynamics

The specific heat of a gas at constant pressure as compared to that at constant volume is

  1. less

  2. equal

  3. more

  4. constant


Correct Option: C
Explanation:

When the gas is heated at constant pressure, some amount of heat is used up in increasing the volume of the gas. For a constant volume process no such heat is required. Thus $C _p>C _v$

The molar specific heat of an ideal gas at constant pressure and volume are $C _p$ and $C _v$ respectively. The value of $C _v$ is

  1. $R$

  2. $\gamma$ R

  3. $\dfrac{R}{\gamma-1}$

  4. $\dfrac{\gamma R}{\gamma-1}$


Correct Option: C
Explanation:

We know that $\displaystyle\dfrac{C _p}{C _v}=\gamma$ and $C _p-C _v=R$.
Thus we get $C _v(\displaystyle\dfrac{C _p}{C _v}-1)=R$
or, $C _v=\displaystyle\dfrac{R}{\gamma -1}$

The gaseous mixture consists of $16\quad $ of helium and $16\quad $ of oxygen. The ratio $\cfrac { { C } _{ p } }{ { C } _{ v } } $ of the mixture is :-

  1. 1.59

  2. 1.62

  3. 1.4

  4. 1.54


Correct Option: B

Calculate the specific heat of a gas at constant volume from the following data. Density of the gas at N.T.P =$19 \times 10 ^ { - 2 } \mathrm { kg } / \mathrm { m } ^ { 3 }$ $\left( C _ { p } / C _ { v } \right)$ = 1.4,J =$4.2 \times 10 ^ { 3 } \mathrm { J } / \mathrm { kcal }$ atmospheric pressure=$1.013 \times 10 ^ { 5 } N / m ^ { 2 }$ (in kcal /kg k)

  1. $2.162$

  2. $1.612$

  3. $1.192$

  4. $2.612$


Correct Option: A

The ratio of the specific heat of air at constant pressure to its specific heat constant volume is

  1. Zero

  2. Greater than one

  3. Less than one

  4. Equal to one


Correct Option: B
Explanation:

The correct answer is option(B).

The ratio of specific heat at constant pressure to the specific heat at constant volume is always greater than one.
As, when the gas is allowed to expand resulting in constant pressure, some of the heat is converted to work resulting in the need of a higher amount of heat to raise the temperature of the gas. Whereas when the volume of the gas is constant, the entire heat supplied is utilized in raising the gas temperature. Hence the heat required the raise the temperature of a unit mass of gas at constant pressure is greater than that required at constant volume. Hence the ratio $c _p:c _v$ is always greater than one.

Which of the following formula is wrong?

  1. $\displaystyle{C _{v} = \dfrac{R}{\gamma - 1}}$

  2. $\displaystyle{C _{p} = \dfrac{\gamma R}{\gamma - 1}}$

  3. $\displaystyle \dfrac{C _{p}}{ C _{v}} = \gamma$

  4. $C _{p} - C _{v} = 2R$


Correct Option: D
Explanation:

The different formula for specific heats is given by:

  • $\dfrac{C _{p}}{C _{v}} = \gamma$
  • $C _{p} - C _{v} = R$
Upon further simplification, we get:
  • $C _{p} = \dfrac{\gamma R}{\gamma -1}$
  • $C _{v} = \dfrac{R}{\gamma -1}$
The incorrect formula is
$C _{p} - C _{v} = 2R$
Hence option D is the answer.

For a gas the ratio of the two specific heats is $\dfrac{5}{3}$. If R $=$ 2 cal /mol-K then the values of $C _{p}$ and $C _{v}$ in cal / mol- K 

  1. $C _p=5 ,C _v=3 $

  2. $C _p=3 ,C _v=4 $

  3. $C _p=4 ,C _v=3 $

  4. $C _p=3 ,C _v=5 $


Correct Option: A
Explanation:

From given data we have $C _{p} - C _{v} = 2 $
and $\dfrac{C _{p}} { C _{v}} = \dfrac {5}{3} $
Solving both gives , 
$C _{p} = 5$ and  $ C _{v} = 3 $

A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. Then $\dfrac{C _{p}}{C _{v}}$ is

  1. 1.67

  2. 2.14

  3. 1.29

  4. 1.33


Correct Option: D
Explanation:

The diatomic molecule has total 6 degress of freedom(3 translational, 2 rotational and 1 vibrational)
Now $C _p$ is given as $(1+\dfrac{f}{2})R=(1+\dfrac{6}{2})R=4R$
and $C _v$ is given as $\dfrac{f}{2}R=\dfrac{6}{2}R=3R$
Thus we get $\dfrac{C _p}{C _v}=\dfrac{4}{3}=1.33$

Which of the following formula is wrong ?

  1. $C _{v}=\dfrac{R}{\gamma -1} $

  2. $\dfrac{C _{p}}{C _{v}}=\gamma $

  3. $C _{p}=\dfrac{\gamma R}{\gamma -1} $

  4. $C _{p}-C _{v}=2R $


Correct Option: D
Explanation:

C$ _p$-C$ _v$ = R and hence option D is incorrect

If the ratio of sp.heat of a gas at constant pressure to that at constant volume is $\gamma $ , the change in internal energy of gas, when the volume changes from V to 2V at constant pressure P is 

  1. $\dfrac{R}{\gamma -1}$

  2. PV

  3. $\dfrac{PV}{\gamma -1}$

  4. $\dfrac{\gamma PV}{\gamma -1}$


Correct Option: C
Explanation:

The change in internal energy in the process should have been,
U = $ nC _v \Delta T $
Now, for this process, if $ \dfrac{{C} _{p}}{{C} _{v}} = \gamma $ and $C _p-C _v=R$
Then, $C _v =  \dfrac{R}{\gamma - 1} $
U = $ \dfrac{nR \Delta T}{\gamma - 1} $
Now, $ nR \Delta T = P(2V - V) $
Thus, U = $ \dfrac{PV}{\gamma - 1} $