Tag: work done by an ideal gas in isothermal expansion

Questions Related to work done by an ideal gas in isothermal expansion

For an isothermal expansion of an ideal gas, mark wrong statement

physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

  1. there is no change in the temperature of the gas

  2. there is no change in the internal energy of the gas

  3. the work done by the gas is equal to the heat supplied to the gas

  4. the work done by the gas is equal to the change in its internal energy

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Correct Option: D
Explanation:

$In\quad isothermal\quad expansion.\ The\quad temperature\quad is\quad constant,\ so\quad the\quad change\quad in\quad temperature,\quad \triangle T\quad =\quad 0,\ as\quad \triangle v\quad =\quad \frac { 3 }{ 2 } \quad R& T\ and\quad \triangle T\quad =\quad 0\ \therefore \quad \quad \triangle V\quad =\quad 0\ \ \quad \quad \quad \quad \quad \quad \quad \triangle V\quad =\quad Q-W\ \quad \quad \quad \quad \quad \quad \quad \quad 0\quad =\quad Q-W\ \quad \quad \quad \quad \quad \quad \quad \quad Q\quad =\quad W\ Work\quad done\quad in\quad isothermal\quad process\quad =\quad nRln\frac { { V } _{ 2 } }{ { V } _{ 1 } } \ \quad \quad \quad nRln\frac { { V } _{ 2 } }{ { V } _{ 1 } } \quad \neq \quad 0\quad \ \quad \therefore \quad \triangle W\quad \neq \quad \triangle V\ Answer\quad :\quad D$

If a given mass of gas occupies a volume of 10 cc at 1 atmospheric pressure and temperature 100$^o$C. What will be its volume at 4 atmospheric pressure, the temperature being the same?

physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

  1. 100 cc

  2. 400 cc

  3. 1.04 cc

  4. 2.5 cc

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Correct Option: D
Explanation:

It is an isothermal process.
$P _1 V _1 = P _2 V _2$
$1 \times 10 = 4 \times V _2$
$V _2 = 2.5 cc$

The cyclic process from X to Y is an isothermal process.
If the pressure of the gas at X is 4.0 kPa, and the volume is 6.0 cubic meters, and if the pressure at Y is 8.0 kPa, what is the volume of the gas at Y?

physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

  1. 12.0 cubic meters

  2. 16.0 cubic meters

  3. 3.0 cubic meters

  4. 4.0 cubic meters

  5. 2.0 cubic meters

Show answer
Correct Option: C
Explanation:
Given :    $P _x = 4.0$ kPa          $V _x = 6.0$  $m^3$                         $P _y = 8.0$ kPa 
X to Y is an isothermal process, thus temperature remains constant   i.e     $T = constant$  or   $PV = constant$
$\implies$     $P _yV _y = P _xV _x$
OR       $8.0 \times V _y  = 4.0 \times 6.0$                     $\implies V _y = 3.0$  $m^3$

The work done y a gas is an isothermal change where 1 refers to initial state and 2 refers to final state is

physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

  1. $\mu R({T _2} - {T _1})\ln \left( {\frac{{{V _2}}}{{{V _1}}}} \right)$

  2. $\mu R{T _1}\ln \left( {\frac{{{V _2}}}{{{V _1}}}} \right)$

  3. $\mu R{T _2}\ln \left( {\frac{{{V _1}}}{{{V _2}}}} \right)$

  4. $\mu R\left( {\frac{{{T _1} + {T _2}}}{2}} \right)[\ln {V _2} - \ln {V _1}]$

Show answer
Correct Option: B
Explanation:

$w=-\int PdV   PV=nRT$
$T\rightarrow const$
$=-\int _{v _1}^{v _2}\frac {nRT}{V}dV$
$=\mu RT _1ln(\frac {V _2}{V _1})$

A thin tube of uniform cross-section is sealed at both ends. When it lies horizontally, the middle $5$cm length contains mercury and the two equal ends contain air at the same pressure P. When the tube is held at an angle of $60^o$ with the vertical, then the lengths of the air columns above and below the mercury column are $46$cm and $44.5$cm respectively. Calculate the pressure P in cm of mercury. The temperature of the system is kept at $30^o$C.

physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

  1. $75.4$

  2. $45.8$

  3. $67.5$

  4. $89.3$

Show answer
Correct Option: A
Explanation:

Let A be the area of cross-section of the tube. When the tube is horizontal, the $5$cm column of Hg is in the middle, so length of air column on either side at pressure $P=\dfrac{46+44.5}{2}=45.25$cm
When the tube is held at $60^o$ with the vertical, the lengths of air columns at the bottom and the top are $44.5$cm and $46$cm respectively. If $P _1$ and $P _2$ are their pressures, then $P _1-P _2=5\cos 60^o=5\times \dfrac{1}{2}=\dfrac{5}{2}$cm of Hg
Using Boyle's law for constant temperature,
$PV=P _1V _1=P _2V _2$
$P\times A\times 45.25=P _1\times A\times 44.5=P _2\times A\times 46$
$\therefore \dfrac{P\times 45.25}{44.5}-\dfrac{P\times 45.25}{46}=\dfrac{5}{2}$
or $P=\dfrac{5\times 44.5\times 46}{2\times 45.25\times 1.5}=75.4$cm of Hg

An ideal gas system undergoes an isothermal process, then the work done during the process is:

physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

  1. $nRT ln\dfrac { { V } _{ 2 }}{ { V } _{ 1 } }$

  2. $nRT ln\dfrac { { V } _{ 1 }}{ { V } _{ 2 }}$

  3. $2nRT ln\dfrac { { V } _{ 2 }}{ { V } _{ 1 }}$

  4. $2nRT ln\dfrac { { V } _{ 1 }}{ { V } _{ 2 }}$

Show answer
Correct Option: A
Explanation:
Let an ideal gas goes isothermally from its initial state $(P _1V _1)$ to the final state $(P _2V _2)$. Then the work done,

$W=\int _{V _1}^{V _2}P\,dV$

We have 

$PV=nRT$

or

$P=\dfrac{nRT}{V}$

Then,

$W=\int _{V _1}^{V _2} \dfrac{nRT}{V} dV=nRT\int _{V _1}^{V _2}\dfrac 1V dV$

$=nRT \,ln(V _2)-ln(V _1)=nRT\,ln\dfrac{V _2}{V _1}$

A vertical cyclinder with heat - conducting with heat conducting walls is closed at the bottom and its fitted with a smooth light piston. It contains one mole of an ideal gas. The temperature of the gas is always equal to the surrounding's temperature $T _o$ . The piston is moved up slowly to increase the volume of the gas to $n$ times. Which of the following is incorrect?

physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

  1. Work done by the gas in $RT _o\ln (n)$.

  2. Work done against the atmosphere is $RT _o(n-1).$

  3. There is no change in the internal energy of the gas.

  4. The final presure of the gas is $\dfrac{1}{n-1}$ times its initial pressure.

Show answer
Correct Option: D
Explanation:

Work Done by gas (isothermal process) : 

$W=nRT\ln(\dfrac { { V } _{ 2 } }{ { V } _{ 1 } } )=R{ T } _{ o }\ln(n)$
Work Done against Atmosphere (Pressure of atmosphere is constant P & Volume is decreased): 

$W=P\triangle V\quad =P(nV-V)=PV(n-1)=R{ T } _{ o }(n-1)$

As there is no change in temperature hence no change is Internal Energy (Always in synchronization with atmosphere).
 Ideal Gas Equation:
$PV=nRT\ PV=R{ T } _{ o }\ P'(nV)=R{ T } _{ o }\ P'=\dfrac{P}{n}$

Two moles of a gas is expanded to double its volume by two different processors. One is isobaric and the other is isothermal. If ${w} _{1}$ and ${w} _{2}$ are the works done respectively, then

physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

  1. ${ w } _{ 2 }=\cfrac { { w } _{ 1 } }{ \ln { 2 } } $

  2. ${ w } _{ 2 }={ w } _{ 1 }$

  3. ${ w } _{ 2 }={ w } _{ 1 }\ln { 2 } $

  4. ${ w } _{ 1 }^{ 2 }={ w } _{ 2 }\ln { 2 } $

Show answer
Correct Option: C
Explanation:

Volume is changed from  $V _1 = V _o$  to  $V _2 = 2V _o$.
Temperature and pressure of gas at $V _o$ are  $T _o$ and $P _o$ respectively.
Thus, from ideal gas equation   $P _o V _o =nRT _o$           .....(1)
Work done in isobaric process  $w _1 = P _o (V _2-V _1) = P _o (2V _o - V _o) = P _o V _o$       .....(2)
Work done in isothermal process  $w _2 = nRT _o \ln\dfrac{V _2}{V _1}$
$\therefore$   $w _2 = nRT _o \ln \dfrac{2V _o }{V _o} = nRT _o \ln 2$
Or   $w _2 = P _oV _o \ln 2$   (from 1)
$\implies \ w _2 = w _1\ln 2$  

Let $ \triangle W _a and \triangle W _b $ the work done by the system A and B respectively in the previous question 

physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

  1. $ \triangle W _a > \triangle W _b $

  2. $ \triangle W _a = \triangle _b $

  3. $ \triangle W _a < \triangle W _b $

  4. The relation between $ \triangle W _a and \triangle W _b $ cannot be deduced.

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Correct Option: C