Tag: heat and thermodynamics

Quizzes (20) Questions (162)

Questions Related to heat and thermodynamics

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

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:

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

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

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

  4. $C _{p}-C _{v}=R$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.
$\int C _PdT=\int C _VdT+P\Delta V$
$\implies C _P<C _V$

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

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 $R=8.31 J/ mo 1.K)$

  1. $62 J$

  2. $104 J$

  3. $124 J$

  4. $208 J$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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

  1. $8\ J/K$

  2. $0.8\ J/K$

  3. $4.0\ J/K$

  4. $0.4\ J/K$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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)

  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} $
Reveal answer Fill a bubble to check yourself
B Correct answer
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, $M _{C _p}-M _{C _v}=R$

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

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

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

  1. $7:5$

  2. $5:7$

  3. $5/2 :7/2$

  4. $3:2$
Reveal answer Fill a bubble to check yourself
C Correct answer
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, $\Delta U:\Delta Q=5/2:7/2$

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

$310 J$ 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

  1. $452J$

  2. $276J$

  3. $144J$

  4. $384J$
Reveal answer Fill a bubble to check yourself
C Correct answer
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\\ $
$Q=n{ C } _{ V }\triangle T\\ =2\times 7.186\times 10\\ =143.72J\approx 144J$
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

$C _p$ 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 :

  1. $a=28 b$

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

  3. $a=b$

  4. $a=14b$
Reveal answer Fill a bubble to check yourself
D Correct answer
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)
  $\implies  a = 14b$
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A gaseous mixture consists of $16\ g$ of helium and $16\ g$ of oxygen, then the ratio $\dfrac { { C } _{ p } }{ { C } _{ v } } $of the mixture is

  1. $1.4$

  2. $1.54$

  3. $1.59$

  4. $1.62$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Hellium is monoatomic.

Oxygen is diatomic.
Degree of freedom of He, ${ f } _{ 1 }=$3
Degree of freedom of ${ O } _{ 2 }$, ${ f } _{ 2 }=$5
$\therefore \cfrac { { C } _{ p } }{ { C } _{ v } }$ of mixture.
$\cfrac { { C } _{ p } }{ { C } _{ v } } =\cfrac { [{ N } _{ 1 }(2+{ f } _{ 1 })+{ N } _{ 2 }(2+{ f } _{ 2 })] }{ { N } _{ 1 }{ f } _{ 1 }+{ N } _{ 2 }{ f } _{ 2 } }$
Putting,${ N } _{ 1 }={ N } _{ 2 }=16\quad gm,\quad$ and $\ { f } _{ 1 }=3,\quad { f } _{ 2 }=5$
we get,$ \cfrac { { C } _{ p } }{ { C } _{ v } } =\cfrac { 3 }{ 2 } \ =1.62$

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

When a heat of Q is supplied to one mole of a monatomic gas $\left ( \gamma =5/3 \right )$, the molar heat capacity of the gas at constant volume is

  1. $ \dfrac{3R}{4}$

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

  3. $ \dfrac{7R}{4}$

  4. $\dfrac{3R}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given $\gamma =5/3$

i.e $\displaystyle \dfrac {C _p}{C _v}=\dfrac {5}{3}$
and $C _p-C _v=R$

$\therefore \displaystyle \dfrac {C _v+R}{C _v}=\dfrac {5}{3}$
$\displaystyle 1+\dfrac {R}{C _v}=\dfrac {5}{3}$

$C _v= \dfrac{3R}{2}$
Option D.