Tag: isothermal and adiabatic processes

Questions Related to isothermal and adiabatic processes

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

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

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

  2. $Q _4 \times Q _T$

  3. $Q _4 + Q _T$

  4. $Q _4 - Q _T$

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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

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

  1. a $=$ 16b

  2. b $=$ 16a

  3. a $=$ 14b

  4. a $=$ b

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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:

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

  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 } $

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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

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

  1. $a=16b$

  2. $b=16a$

  3. $a=4b$

  4. $a =b$

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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.

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

  1. $0.287\ KJ/kg K$

  2. $0.21\ kJ/kg K$

  3. $0.34\ kJ/kg K$

  4. $0.19\ kJ/kg K$

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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$  $=$8.314 kJ/Kmole K. (Molecular weight of argon$=$ $39.95$)

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

  1. $520.3$

  2. $530.2$

  3. $230.5$

  4. $302.5$

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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)

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

  1. 28 cal

  2. 56 cal

  3. 84 cal

  4. 94 cal

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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.

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

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

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

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

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

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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)

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

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

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

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

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

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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=$\dfrac{0.0397333}{6.02\times 10^{23}}=6.60\times 10^{-23}g$

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

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

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

  2. $PV$

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

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

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Correct Option: C
Explanation:

At constant pressure, change in internal energy$ \Delta U=nC _v\Delta T$
Now,$\dfrac{C _p}{C _v}=\gamma$
$\Rightarrow 1$+$\dfrac{R}{C _v}$=$\gamma$
$\Rightarrow C _v=\dfrac{R}{\gamma -1}$
Using Charle's law, final temperature=2\times initial temperatue=2T
Thus,  $ \Delta U=nC _v(2T-T)=nC _vT=\dfrac{nRT}{\gamma -1}=\dfrac{PV}{\gamma -1}$