Tag: heat and thermodynamics

Questions Related to heat and thermodynamics

In an isobaric process, the correct ratio is :

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

  1. $\Delta Q:\Delta W=1:1$

  2. $\Delta Q:\Delta W=\gamma :\gamma -1$

  3. $\Delta Q:\Delta W= \gamma -1:\gamma $

  4. $\Delta Q:\Delta W= \gamma :1 $

Show answer
Correct Option: B
Explanation:

In an isobaric process, pressure is constant.
Heat added in an isobaric process is given by
$\Delta Q=\Delta U+ \Delta W$
$nC _p\Delta T=nC _v\Delta T+nR\Delta T$
$\displaystyle \therefore \dfrac {\Delta Q}{\Delta W}=\dfrac {nC _p\Delta T}{nR\Delta T}=\dfrac {C _p}{R}=\dfrac {\gamma R/\gamma -1}{R}=\dfrac {\gamma }{\gamma -1}$
Option B.

A cylinder of fixed capacity $67.2$ liters contains helium gas at STP. Calculate the amount of heat required to raise the temperature of the gas by $15^{o}C$. ($R=8.314\ J\ mol^{-1}k^{-1}$)

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

  1. $520\ J$

  2. $560. J$

  3. $620\ J$

  4. $621.2\ J$

Show answer
Correct Option: B
Explanation:

Since, the process is at constant volume,
Q = U as W = 0
Thus, Q = $ n {C} _{v} \Delta T $
At STP, n = $ \dfrac{PV}{RT} $
Since, He is diatomic, $ {C} _{v} = 2.5R $
Q = $ \dfrac{PV}{RT} \times 2.5R \times 15 $
Substituting the pressure and temperature values at STP, 
P = 1 atm
V = 67.2 L
T = 298 K
we get,
Q = 560.9 J

A diatomic gas is heated at constant pressure. The fraction of the heat energy used to increase the internal energy is 

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

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

  2. $ \dfrac{3}{7}$

  3. $ \dfrac {5}{7}$

  4. $ \dfrac {7}{9}$

Show answer
Correct Option: C
Explanation:

In a diatomic gas, we have $C _p=\dfrac{7}{2}R $ and $C _v=\dfrac{5}{2}R$
The heat is given as $nC _p\Delta T$ and internal energy as $nC _v\Delta T$
Thus we get $\dfrac{U}{Q}$ as $\dfrac{5}{7}$

Four students found set of $C _{p}$ and $C _{v}$[in cal/deg mole] as given below, which of the following set is correct 

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

  1. $C _{v}=4,C _{p}=2$

  2. $C _{v}=4,C _{p}=3$

  3. $C _{v}=3,C _{p}=4$

  4. $C _{p}=5,C _{v}=3$

Show answer
Correct Option: D
Explanation:

C$ _{v}$ cannot be greater than $C _{p}$

Hence, option A  and option B are incorrect.

We have the relation $C _{p} - C _{v}$= R ( and its value is 2 cal/mole ) and hence option C is also incorrect.

A solid copper sphere(density $\rho$ and specific heat c) of radius r at an initial temperature $200$K is suspended inside a chamber whose walls are at almost $0$ K. The time required to the temperature of sphere to drop to $100$ K is _________?

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

  1. $\dfrac{9r\rho c}{72\times 10^6\sigma}$sec.

  2. $\dfrac{7r\rho c}{72\times 10^6\sigma}$sec.

  3. $\dfrac{7r\rho c}{82\times 10^6\sigma}$sec.

  4. $\dfrac{19r\rho c}{72\times 10^7\sigma}$sec.

Show answer
Correct Option: B

If $C _p$ and $C _v$ denote the specific heats (per unit mass) of an ideal gas of molecular weight M, where 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. $C _p - C _v = R/M^2$

  2. $C _p - C _v = R$

  3. $C _p - C _v = R/M$

  4. $C _p - C _v = M/R$

Show answer
Correct Option: C
Explanation:

By definition 

$dU={ C } _{ v }dT\longrightarrow 1$

also enthalpy,

$H=U+PV\\ or\quad dH=dU+d\left( PV \right) \\ or\quad dH=dU+nRdT\longrightarrow 2$

Also $dH={ C } _{ P }dT\\ \therefore { C } _{ P }dT={ C } _{ V }dT+nRdT\\ \Rightarrow { C } _{ P }={ C } _{ V }+nR\\ or{ C } _{ P }-{ C } _{ V }\quad =nR=\cfrac { Rm }{ M } $

for $m=1$

${ C } _{ P }-{ C } _{ V }=\cfrac { R }{ M } $

${C} _{P}$ 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 then

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

  1. $a=28b$

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

  3. $a=b$

  4. $a=14b$

Show answer
Correct Option: D
Explanation:

For ideal gas
${C} _{P}-{C} _{V}=R/M$
If ${C} _{P}$ and ${C} _{V}$ are specific heats $\left( J/kg- _{  }^{ o }{ C } \right) $
$M=$ molar mass of gas
$\Rightarrow a=R/2$ and $b=R/28$
$\Rightarrow$ $a=14b$

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 $30g$)

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

  1. $2.0kcal/kg$

  2. $7kcal/g$

  3. $3kcal/kg$

  4. $0.5kcal/kg$

Show answer
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
$\displaystyle \frac{Heat  Utilised}{Heat  Produced}$

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. $Q _4 - Q _T$

Show answer
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

Show answer
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.