Tag: work done by an ideal gas in isothermal expansion

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

Identical cylinders contain helium at 2.5 atm and agron at 1 atm respectively. If the are filled  in one of the cylinder,the pressure would be.

  1. 3.6 atm

  2. 1.75 atm

  3. 1.5 atm

  4. 1.0 atm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given that,

Pressure ${{p} _{1}}=2.5\,atm$

Pressure ${{P} _{2}}=1\,atm$

Volume ${{V} _{1}}={{V} _{2}}=V$

Both the cylinders are similar, volume of both the gases is equal

Let the volume of gases be V

The amount of pressure P in one of the cylinder will be equal to the total pressure at equilibrium

Now,

  $ P=\dfrac{{{P} _{1}}{{V} _{1}}+{{P} _{2}}{{V} _{2}}}{{{V} _{1}}+{{V} _{2}}} $

 $ P=\dfrac{2.5\times V+1\times V}{2V} $

 $ P=1.75\ atm $

Hence, the pressure is $1.75\ atm$

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

One mole of an ideal monoatomic gas is at $360K$ and a pressure of $10 ^ { 5 } Pa.$ It is compressed at constant pressure until its volume is halved. Taking $R$ as $8.3 J{ mol } ^ { - 1 }{ K } ^ { - 1 }$ and the initial volume of the gas as $3.0 \times 10 ^ { - 2 } { m } ^ { 3 }$ , the work done on the gas is

  1. $-1500 J$

  2. $+1500 J$

  3. $-3000 J$

  4. $+3000 J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes 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.

  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.
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
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.

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

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?

  1. $2$

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

  3. $\ln 2$

  4. $\ln 3$
Reveal answer Fill a bubble to check yourself
C Correct answer
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=$\dfrac{PV \ln2}{PV}=ln2$

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

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

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The work done in a reversible isothermal process for an ideal gas is derived from the integral of PdV, resulting in W = nRT ln(V2/V1). Using base-10 logarithms, this is 2.303 nRT log(V2/V1).

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

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

  1. $2\ J$

  2. $200\ J$

  3. $300\ J$

  4. $2 \times 10^{5}\ J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work done = P * deltaV. P = 1 atm = 1.013 * 10^5 Pa. deltaV = 3L - 1L = 2L = 2 * 10^-3 m^3. W = 1.013 * 10^5 * 2 * 10^-3 = 202.6 J, which is approximately 200 J.

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

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

  1. 155 J

  2. 255 J

  3. 355 J

  4. 555 J
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

During the process A -B of an ideal gas: 

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

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).

  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. $\dfrac {3P _{0}V _{0}}{R}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given PV = RT + 3V. For isobaric expansion, P is constant. Differentiating: P dV = R dT. Work done W = P * deltaV = R * deltaT. Since temperature doubles, deltaT = T0. Thus W = R * T0. The provided option A seems to be a complex derivation involving the specific equation of state.