Work done by an ideal gas in isothermal expansion - class-XI

Work done is the area under the P-V curve. For a given expansion, the adiabatic curve is the steepest, resulting in the smallest area under the curve compared to isothermal or isobaric processes.

$w=nRT ln\left(\dfrac{v _2}{v _1}\right)$
$w=P _0V _0ln\left(\dfrac{v _2}{v _1}\right)$
$w=10^5\times 10\times 10^{-3}ln\left(\dfrac{1}{15}\right)$
$w=-2.70\times 10^3J$.

Under isothermal process work is given by the relation $W = RT \ln (\dfrac{V _{f}}{V _{i}})$
Therefore work will be $W = RT \ln(2)$

The slope of an adiabatic process is given by -gamma * P/V, while the slope of an isothermal process is -P/V. Since gamma (Cp/Cv) is always greater than 1 for an ideal gas, the adiabatic slope is steeper (greater in magnitude).

The process involves isothermal expansion, isobaric compression, and isochoric heating. Calculating work for each step and summing them yields the net work; for a cycle, the heat supplied equals the work done if the internal energy change is zero, but here we sum the heat for each leg.

In an isothermal process for an ideal gas, the change in internal energy (dU) is zero because temperature is constant. According to the first law of thermodynamics (dQ = dU + dW), dQ must equal dW.

Work done in isothermal expansion is W = nRT ln(P1/P2). Using n=1, R=8.314 J/molK, T=300K, and ln(10/1) = 2.303, we get W = 1 * 8.314 * 300 * 2.303 = 5744 J. Since it is expansion, the gas does work, but the question asks for work done during expansion (often defined as positive by convention in physics).

Work done on the gas during compression at constant pressure is given by W = -P deltaV. Since the volume is halved, deltaV = V_final - V_initial = (3.0 x 10^-2 / 2) - 3.0 x 10^-2 = -1.5 x 10^-2 m^3. Thus, W = -10^5 * (-1.5 x 10^-2) = +1500 J, which is the work done on the gas.

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)

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=$\dfrac{PV}{2V}=P/2$. 

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).

When two ideal gases are mixed isothermally, the maximum work done corresponds to the sum of the work of expansion for each gas into the total volume. For 0.5 moles of oxygen and 0.5 moles of nitrogen (total 1 mole of gas) expanding from volume V to 2V, the work is n R T ln(V_final / V_initial) = 1 * R * T * ln(2) = RT ln 2.

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.

Given,

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.

One calorie is equivalent to $4.184$ joules 

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.

$\triangle Q = \left (\dfrac {f}{2} + 1\right ) nR\triangle T = \dfrac {7}{2} \times 10 = 35\ J$.

For an ideal gas, internal energy E is a function of temperature only (E = f(T)). In an isothermal process, deltaT = 0, therefore deltaE = 0.

$\dfrac{f}{2} R = \dfrac{C _v}{n}$
$W = nR \Delta T$
$\Delta Q = \left(\dfrac{f}{2} + 1\right) nR , \Delta T$
$\dfrac{W}{\Delta Q} = \left(\dfrac{2}{f + 2} \right) = \dfrac{2}{\dfrac{2C _v}{nR} + 2} = \left(\dfrac{nR}{C _v + R} \right)$

Work done in an isothermal process is given by 
$\displaystyle{W = 2.3026nRT\log _{10}\frac{V _2}{V _1}}$
Here, $n = 3, R = 8.31 J/mol^oC$
$T = 300 K$, $V _1= 4 L$, $V _2 = 1 L$
Hence, $\displaystyle{W = 2.3026 \times 3 \times 8.31 \times 300 \times log _{10}\frac{1}{4}}$
= $17221.15 (-2\log _{10} 2$)
= $-17221.15 \times 2 \times 0.3010 = -10368J$ 

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

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

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

Work Done by gas (isothermal process) : 

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)
$\implies \ w _2 = w _1\ln 2$