Tag: principal and molar specific heats of gases

Questions Related to principal and molar specific heats of gases

A mass of $50g$ of water in a closed vessel with surroundings at a constant temperature takes $2$ minutes to cool from ${30}^{o}C$ to ${25}^{o}C$. A mass of $100g$ of another liquid in an identical vessel with identical surroundings takes the same time to cool from ${30}^{o}C$ to ${25}^{o}C$. The specific heat of the liquid is : (The water equivalent of the vessel is $30g$)

  1. $2.0kcal/kg$

  2. $7kcal/g$

  3. $3kcal/kg$

  4. $0.5kcal/kg$


Correct Option: D
Explanation:
As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from $30^{\circ} C$ to $25^{\circ} C$) is same 2 minutes.

$\therefore \left(\dfrac{dQ}{dt} \right) _{water} = \left(\dfrac{dQ}{dt} \right) _{liquid}$

Or $\dfrac{(m _w c _w + W) \Delta T _1}{t _1}  = \dfrac{(m _l c _l + W) \Delta T _2}{t _2}$

$\therefore \Delta T _1=\Delta T _2, \, t _1=t _2$

(w = water equivalent of the vessel)

or $m _w c _w = m _l c _l$

$\therefore$ specific heat of liquid,

$C _l = \dfrac{m _w c _w}{m _l} = \dfrac{50 \times 1}{100} = 0.5 kcal/kg$

Thermal efficiency $=$ .........................   or
$\displaystyle \frac{Heat  Utilised}{Heat  Produced}$

  1. $\displaystyle \frac{Q _4}{Q _T}$

  2. $Q _4 \times Q _T$

  3. $Q _4 + Q _T$

  4. $Q _4 - Q _T$


Correct Option: A

For hydrogen gas $C _{p}-C _{v}=a$ and for Oxygen gas $C _{p}-C _{v}=b $, where $C _{p}$ and $C _{v}$ are molar specific heats. Then the relation between a and b. is

  1. a $=$ 16b

  2. b $=$ 16a

  3. a $=$ 14b

  4. a $=$ b


Correct Option: D
Explanation:

For any ideal gas,$C _p-C _v=nR$, where $R$ is the gas constant.
That is $C _p-C _v$ per mole for any gas is a constant value.
So, $a=b$
Option D.

Three perfect gases at absolute temperatures ${T} _{1},{T} _{2}$ and ${T} _{3}$ are mixed. The masses of molecules are ${m} _{1},{m} _{2}$ and ${m} _{3}$ and the number of molecules are ${n} _{1},{n} _{2}$ and ${n} _{3}$ respectively. Assuming no loss of energy, the final temperature of the mixture is:

  1. $\cfrac { { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ T } _{ 2 }+{ n } _{ 3 }{ T } _{ 3 } }{ { n } _{ 1 }+{ n } _{ 2 }+{ n } _{ 3 } } $

  2. $\cfrac { { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ { T } _{ 2 } }^{ 2 }+{ n } _{ 3 }{ { T } _{ 3 } }^{ 2 } }{ { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ T } _{ 2 }+{ n } _{ 3 }{ T } _{ 3 } } $

  3. $\cfrac { { n } _{ 1 }{ { T } _{ 1 } }^{ 2 }+{ n } _{ 2 }{ { T } _{ 2 } }^{ 2 }+{ n } _{ 3 }{ { T } _{ 3 } }^{ 2 } }{ { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ T } _{ 2 }+{ n } _{ 3 }{ T } _{ 3 } } $

  4. $\cfrac { \left( { T } _{ 1 }+{ T } _{ 2 }+{ T } _{ 3 } \right) }{ 3 } $


Correct Option: A

For hydrogen gas $C _{p} -C _{v} = a$ and for oxygen gas $C _{p} - C _{v}=b$, where $C _{p}$ and $C _{v}$ are molar specific heats. Then the relation between 'a' and 'b' is

  1. $a=16b$

  2. $b=16a$

  3. $a=4b$

  4. $a =b$


Correct Option: A

The specific heat of air at constant pressure is $1.005\ kJ/kg\ K$ and the specific heat of air at constant volume is $0.718\ kJ/kg\ K$ .Find the specific gas constant.

  1. $0.287\ KJ/kg K$

  2. $0.21\ kJ/kg K$

  3. $0.34\ kJ/kg K$

  4. $0.19\ kJ/kg K$


Correct Option: A
Explanation:

Specific gas constant = Specific heat at constant pressure - Specific heat at constant volume

                                     = 1.005 - 0.718
                                     = 0.287 KJ/kgK

The specific heat of Argon at constant volume is $0.3122 kj/kg K$. Find the specific heat of Argon at constant pressure if  $ R$  $=$8.314 kJ/Kmole K. (Molecular weight of argon$=$ $39.95$)

  1. $520.3$

  2. $530.2$

  3. $230.5$

  4. $302.5$


Correct Option: A
Explanation:
Given,
$C _v=0.3122\ kJ/kg.K$
$R=8.314$
$M=39.95$
$C _{p}=?$
We know,

$C _p-C _v=\dfrac{R}{M}$

$C _p-C _v=\dfrac{8.314}{39.95}=0.2081$

$C _p=0.3122+0.2081=0.5203$

$C _p=520.3\ J/kg.K$

Option $\textbf A$ is the correct answer

Four moles of a perfect gas heated to increase its temperature by ${2^ \circ }C$ absorbs heat of 40 cal at constant volume. If the same gas is heated at constant pressure the amount of heat supplied is (R$=$ 2 cal/mol K)

  1. 28 cal

  2. 56 cal

  3. 84 cal

  4. 94 cal


Correct Option: B
Explanation:
Heat supplied at constant volume
$Q _v=nC _v\triangle T$
$40=4\times C _v\times 2$
$C _v=5$ cal/mol.K
$C _p=C _v+R=5+2=7cal/mol.k$
$\Rightarrow $ Heat supplied at constant pressure
$Q _p=nC _p\triangle T=4\times 7\times 2$
$Q _p=56cal$

Eight spherical droplets, each of radius $'r'$ of a liquid of density $'\phi'$ and surface tension $'T'$ coalesce to form one big drop. If $'s'$ in the specific heat of the liquid. Then the rise in the temperature of the liquid.

  1. $\dfrac {2T}{3r \rho s}$

  2. $\dfrac {3T}{r \rho s}$

  3. $\dfrac {3T}{2r \rho s}$

  4. $\dfrac {T}{r \rho s}$


Correct Option: C

The specific heat at constant volume for the monatomic argon is $0.075 \ kcal/kg-K$, whereas its gram molecular specific heat is $C _v \ = 2.98 \ cal/mol/K$. The mass of the argon atom is (Avogadro's number $= 6.02 \times 10^{23}$ molecules/mol)

  1. $6.60 \times 10^{-23} g$

  2. $3.30 \times 10^{-23}g$

  3. $2.20 \times 10^{-23}g$

  4. $13.20 \times 10^{-23}g$


Correct Option: A
Explanation:

Mass of one mole of argon =$\dfrac{gram \  molecular \  specific \  heat}{specific\   heat \  at \  constant \  volume}=\dfrac{2.98\times 10^{-3}}{0.075}=0.039733 \ g$


Thus mass of each argon atom=$\dfrac{0.0397333}{6.02\times 10^{23}}=6.60\times 10^{-23}g$