Tag: heat and thermodynamics

Questions Related to heat and thermodynamics

An ideal gas is taken from state $A$ (pressure $P$, volume $V$) to state $B$ (pressure $\displaystyle\frac{P}{2}$, volume $2V$) along a straight line path in the pressure-volume diagram. Select the correct statements from the following.

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

  1. The work done by the gas in the process $A$ to $B$ exceeds the work done that would be done by it if the system were taken from $A$ to $B$ along an isotherm.

  2. In the temperature-volume diagram, the path $AB$ becomes a part of a parabola.

  3. In the pressure-temperature diagram, the path $AB$ becomes a part of hyperbola.

  4. In going from $A$ to $B$, the temperature $T$ of the gas first increases to a maximum value and then decreases.

Show answer
Correct Option: A,B,D
Explanation:

The given process follows a linear P-V relation given by: $ \dfrac{p-P}{v-V}=\dfrac{P-P/2}{V-2V}=-P/2V$
$\Rightarrow p-P=\dfrac{-P}{2V}(v-V) \Rightarrow p=-\dfrac{Pv}{2V}+3P/2$        .......(i)


Here work done $\Delta W$=area under P-V diagram =$3PV/4=0.75PV$
For isotherm, pv=PV $\Rightarrow$ work done $\Delta W$=$PV\ln2=0.693PV$

Hence statement a is correct
For t-v diagram, replace p in (i) by $\dfrac{nRt}{v}$
Hence, $t=-\dfrac{Pv^2}{2nRV}+\dfrac{3Pv}{2nR}$

The above eqn. has t equalling a quadratic in v. 
Hence, t-v diagram is a parabola $\Rightarrow$ statement b is correct.

Clearly the above parabola is concave downwards , hence has a maxima.
The t has same initial and final value implying the maxima occurs in-between.

Hence statement d is correct.

A fixed mass of a gas is first heated isobarically to double the volume and then cooled isochorically to decrease the temperature back to the initial value. By what factor would the work done by the gas decreased, had the process been isothermal?

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

  1. $2$

  2. $\displaystyle\dfrac{1}{2}$

  3. $\ln 2$

  4. $\ln 3$

Show answer
Correct Option: C
Explanation:

Let initial pressure and volume be P  and V respectively.
Then after isobaric expansion they are P and 2V. Here work done=$P\Delta V=P(2V-V)=PV$
To bring to initial temperature new pressure=$\dfrac{PV}{2V}=P/2$. 


So after isochoric process they are $P/2  \ and \  2V$ . Here, $\Delta V=0\Rightarrow$ work done=$P\Delta V=0$
Hence, total work done in the 2 successive processes=PV +0=PV
For isothermal expansion, work done=$PVln(\dfrac{V _f}{V _i})=PV\ln2$
Hence required factor=$\dfrac{PV \ln2}{PV}=ln2$

Work done in reversible isothermal process by an ideal gas is given by

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

  1. $2.303 \text { nRT log } \frac { V _ { 2 } } { V _ { 1 } }$

  2. $\frac { n R } { ( y - 1 ) } \left( T _ { 2 } - T _ { 1 } \right)$

  3. $2.303 \text { nRT log } \frac { V _ { 1 } } { V _ { 2 } }$

  4. None

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Correct Option: A

$0\cdot 5\ mole$ of oxygen and $0\cdot 5\ mole$ of nitrogen, each having $V$ and temperature $T$ are mixed isothermally to have the total volume $2\ V$. The maximum work done is :

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

  1. $RT\ \log _{e}2\ joule$

  2. $\dfrac {RT}{2}\ \log _{e}2\ joule$

  3. $RT\ (\log _{e}4)\ joule$

  4. $zero$

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Correct Option: C

A gas expands from $1l$ to $3l$ at atmospheric pressure. The work done by the gas is about 

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

  1. $2\ J$

  2. $200\ J$

  3. $300\ J$

  4. $2 \times 10^{5}\ J$

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Correct Option: B

4 atm pressure to attain 1 atm pressure by result of isothermal  expansion. The work done by the gas during expansion is nearly.

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

  1. 155 J

  2. 255 J

  3. 355 J

  4. 555 J

Show answer
Correct Option: D
Explanation:

Given,

$V=1L=0.001m^3$
$P=4atm$
$\dfrac{P _1}{P _2}=4$
From ideal gas,
$PV=nRT$
$T=\dfrac{PV}{nR}$
Work done during the isothermal expansion,
$W=nRTln(\dfrac{P _1}{P _2})$
$W=nR.\dfrac{PV}{nR}ln(4)$
$W=PVln(4)$
$W=4\times 10^5\times 0.001\times ln(4)$
$W=555J$
The correct option is D.

During the process A -B of an ideal gas: 

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

  1. work done on the gas is zero

  2. density of the gas is constant

  3. slope of line AB from the T - axis is inversely proportional to the number of moles of the gas

  4. slope of line AB from the T - axis is directly proportional to the number of moles of the gas

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Correct Option: A

A gas follow a general process as $PV RT + 3V$ for $1\ mole$ of gas. If it expands isobarically till temperature is doubled, then the work done by the gas is (initial temperature and pressure are $T _{0}$ and $P _{0}$ respectively).

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

  1. $\dfrac {P _{0}T _{0}R}{(2P _{0} - 3)}$

  2. $\dfrac {P _{0}T _{0}R}{(P _{0} - 3)}$

  3. $\dfrac {P _{0}T _{0}R}{(P _{0}V - 3)}$

  4. $\dfrac {3P _{0}V _{0}}{R}$

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Correct Option: A

The work done by 100 calorie of heat in isothermal expansion of ideal gas is 

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

  1. 418.4J

  2. 4.184J

  3. 41.84J

  4. None

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Correct Option: A
Explanation:

One calorie is equivalent to $4.184$ joules 

or,   $J=4.18 J/cal$.
Work done (W)=$ = 4.18 \times 500 = 2090\,J$
As $1$ calorie is equal to $4.184$ calorie then $100$ calorie is equal to $ = 4.184 \times 100 = 418.4$

Work done during isothermal expansion depends on change in

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

  1. volume

  2. pressure

  3. both (a) and (b)

  4. none of these

Show answer
Correct Option: C
Explanation:

Work done in isothermal expansion is given as $W=nRT\ln(\displaystyle\frac{V _f}{V _i})=nRT\ln(\frac{P _i}{P _f})$. Thus it depends both on pressure and volume.