Tag: isothermal and adiabatic processes

Quizzes (2) Questions (144)

Questions Related to isothermal and adiabatic processes

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

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

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

  2. PV

  3. <span>$\dfrac{PV}{\gamma -1}$</span>

  4. <span>$\dfrac{\gamma PV}{\gamma -1}$</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The change in internal energy in the process should have been,
U = $ nC _v \Delta T $
Now, for this process, if $ \dfrac{{C} _{p}}{{C} _{v}} = \gamma $ and $C _p-C _v=R$
Then, $C _v =  \dfrac{R}{\gamma - 1} $
U = $ \dfrac{nR \Delta T}{\gamma - 1} $
Now, $ nR \Delta T = P(2V - V) $
Thus, U = $ \dfrac{PV}{\gamma - 1} $

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

In an isobaric process, the correct ratio is :

  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 $
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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 $15^{o}C$. ($R=8.314\ J\ mol^{-1}k^{-1}$)

  1. $520\ J$

  2. $560. J$

  3. $620\ J$

  4. $621.2\ J$
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

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

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

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

  4. $ \dfrac {7}{9}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $\dfrac{5}{7}$

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

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

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

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

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

  4. $C _{p}=5,C _{v}=3$
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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 _________?

  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.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This involves Stefan-Boltzmann law for cooling. The time taken to cool from T1 to T2 is derived by integrating the rate of heat loss. The correct coefficient for the given parameters leads to option B.

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, where R is the molar gas constant:

  1. $C _p - C _v = R/M^2$

  2. $C _p - C _v = R$

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

  4. $C _p - C _v = M/R$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 } $

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 then

  1. $a=28b$

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

  3. $a=b$

  4. $a=14b$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $b=R/28$
$\Rightarrow$ $a=14b$

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

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 $30g$)

  1. $2.0kcal/kg$

  2. $7kcal/g$

  3. $3kcal/kg$

  4. $0.5kcal/kg$
Reveal answer Fill a bubble to check yourself
D Correct answer
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,

$C _l = \dfrac{m _w c _w}{m _l} = \dfrac{50 \times 1}{100} = 0.5 kcal/kg$