Tag: isothermal and adiabatic processes

Questions Related to isothermal and adiabatic processes

A monatomic ideal gas expands at constant pressure, with heat Q supplied. The fraction of Q which goes as work done by gas is

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

  1. 1

  2. $\displaystyle{\dfrac{2}{3}}$

  3. $\displaystyle{\dfrac{3}{5}}$

  4. $\displaystyle{\dfrac{2}{5}}$

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Correct Option: D
Explanation:

$Q = nC _p \Delta T$ and $W = P\Delta V = nR\Delta T$
monatomic gas, $\displaystyle{C _p = \dfrac{5R}{2}}$.
$\Rightarrow$$\displaystyle{\dfrac{W}{Q} = \dfrac{2}{5}}$

For a solid with a small expansion coefficient

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

  1. $ C _p - C _v = R $

  2. $ C _p = C _v $

  3. $ C _p $ is slightly greater than $ C _v $

  4. $ C _p $ is slightly less than $ C _v $

Show answer
Correct Option: C

When water is heated from $0^{\circ}C$ to $4^{\circ}C$ and $C _{p}$ and $C _{v}$ are its specific heated at constant pressure and constant volume respectively, then:

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

  1. $C _{p} >C _{v}$

  2. $C _{p}< C _{v}$

  3. $C _{p}=C _{v}$

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

Show answer
Correct Option: B
Explanation:

Water has highest density at $4^{\circ}C$. This changes its properties from other simple fluids.

When water is heated from $0^{\circ}C$ to $4^{\circ}C$, the volume of liquid decreases.
Thus for this transition, $P\Delta V$ is negative.
$\int C _PdT=\int C _VdT+P\Delta V$
$\implies C _P<C _V$

Two moles of ideal helium gas are in a rubber balloon at $30^{o}C$. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to $35^{o}C$. The amount of heat required in raising the temperature is nearly $($take $R=8.31 J/ mo 1.K)$

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

  1. $62 J$

  2. $104 J$

  3. $124 J$

  4. $208 J$

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Correct Option: D
Explanation:

For isobaric process.
$ \Delta Q= n C _{p} \Delta T$
$=2 \times \dfrac{5}{2} R \times (35-30)$
$= 208 \ J$

The temperature of $5\ moles$ of a gas which was held at constant volume was changed from $100^{o}C$ to $120^{o}C$. The change in the internal energy of the gas was found to be $80\ J$, the total heat capacity of the gas at constant volume will be equal to

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

  1. $8\ J/K$

  2. $0.8\ J/K$

  3. $4.0\ J/K$

  4. $0.4\ J/K$

Show answer
Correct Option: C
Explanation:

$dU = nC _v dT$ or, $ 80 = 5 \times C _v(120 - 100)$
$C _v = 4.0\ J/K$

The value of the ratio $C _p/C _v$ for hydrogen is 1.67 at 30 K but decreases to 1.4 at 300 K as more degrees of freedom become active. During this rise in temperature

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

  1. $C _p$ remains constant but $C _v$ increases

  2. $C _p$ decreases by $C _v$ increases

  3. both $C _p$ and $C _v$ decreases by the same amount

  4. both $C _p$ and $C _v$ increases by the same amount

Show answer
Correct Option: D
Explanation:

he value of the ratio $\dfrac{Cp}{Cv} $for hydrogen is 1.67 at 30 K but decreases to 1.4 at 300 K as more degrees of freedom become active. During this rise in temperature both $Cp$ and $Cv$ increases by the same amount
 For an ideal gas, $C _p = C _v + R$. If it is a molecular gas, increasing temperature enables vibrational degrees of freedom, so that $C _v$ increases. Hence $\dfrac{C _p}{C _v} = 1 +\dfrac{ R}{C _v}$ decreases.

If $ {C} _{P}$ and $ {C} _{V}$ denote the specific heats (per unit mass) of an ideal gas of molecular weight M then which of the following relations is true ?
(R is the molar gas constant)

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

  1. ${C} _{P}$ - ${C} _{V} = R$

  2. ${C} _{P}$ - ${C} _{V} = R / M$

  3. ${C} _{P}$ - ${C} _{V} = MR$

  4. ${C} _{P}$ - ${C} _{V}$ = $R /{M}^{2} $

Show answer
Correct Option: B
Explanation:

Let $Cu$ and $Cp$ be molar specific heats of the ideal gas at a 


constant volume and constant pressure, respectively, then

$C _p=M _{c _p}$ and $C _v=M _{c _v}$

Where $C _p$ and $C _v$ are specific heat (per unit mass)

if $C _p$ and $C _v$ are specific heat (for unit mass) of an ideal gas of molecular weight $M$

then specific heat (At constant P) for $M=MC _p$ and 

then specific heat (At constant V) for $M=MC _v$ 

then, $M _{C _p}-M _{C _v}=R$

$\boxed{C _p-C _v=R/M}$

If heat energy $\Delta $ is supplied to an ideal diatomic gas and the increase in internal energy is $\Delta U$, the ratio of $\Delta U:\Delta Q$ is

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

  1. $7:5$

  2. $5:7$

  3. $5/2 :7/2$

  4. $3:2$

Show answer
Correct Option: C
Explanation:

For a diatomic gas, the specific heat at constant pressure $C _p=\frac{7}{2}R$ and the specific heat at constant volume $C _v=\dfrac{5}{2}R$

Thus, $\Delta U=nC _v\Delta T=\dfrac{5}{2}nR\Delta T$ and 
$\Delta Q=nC _p\Delta T=\dfrac{7}{2}nR\Delta T$
Hence, $\Delta U:\Delta Q=5/2:7/2$

$310 J$ of heat is required to raise the temperature of $2$ moles of an ideal gas at constant pressure from $25^0C$ to $35^0C$. The amount of heat energy required to raise the temperature of the gas through the same range at constant volume is

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

  1. $452J$

  2. $276J$

  3. $144J$

  4. $384J$

Show answer
Correct Option: C
Explanation:
Heat = moles(no.) $\times C _P \times \triangle T$
$\Rightarrow 310=2\times { C } _{ P }\times 10\quad \quad [35-25=10]\\ \Rightarrow { C } _{ P }=15.5J/molK\\ $
$\therefore { C } _{ P }-{ C } _{ V }=R\\ \Rightarrow { C } _{ V }={ C } _{ P }-R=15.5-8.314\\ \Rightarrow { C } _{ V }=7.186J/molK\\ $
$Q=n{ C } _{ V }\triangle T\\ =2\times 7.186\times 10\\ =143.72J\approx 144J$

$C _p$ and $C _v$ are specific heats at constant pressure and constant volume respectively. It is observed that
$C _p-C _v=a$ for hydrogen gas
$C _p-C _v=b$ for nitrogen gas
The correct 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. $a=28 b$

  2. $a=\dfrac{1}{14}b$

  3. $a=b$

  4. $a=14b$

Show answer
Correct Option: D
Explanation:
For any gas, $C _p-C _v=R$, which is the gas constant.
Hence be it any gas, hydrogen or nitrogen, its value is same.
Here, for ideal gas, $C _p – C _v = R/M$, where $M$ is the mass of one mole of gas.
Mass of one mole of hydrogen  $M = 2$ g and that of nitrogen  $M = 28$ g 
$\therefore$ $a =C _p -C _v= R/2$  (for hydrogen) 
And $b =C _p - C _v = R/28$  (for nitrogen)
  $\implies  a = 14b$