Tag: heat and thermodynamics

Questions Related to heat and thermodynamics

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


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

  1. $62 J$

  2. $104 J$

  3. $124 J$

  4. $208 J$


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

  1. $8\ J/K$

  2. $0.8\ J/K$

  3. $4.0\ J/K$

  4. $0.4\ J/K$


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

  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


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)

  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} $


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

  1. $7:5$

  2. $5:7$

  3. $5/2 :7/2$

  4. $3:2$


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

  1. $452J$

  2. $276J$

  3. $144J$

  4. $384J$


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 :

  1. $a=28 b$

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

  3. $a=b$

  4. $a=14b$


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)
  $\implies  a = 14b$

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. $1.62$


Correct Option: D
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$

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


Correct Option: D
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.