One mole of ideal gas taken through a cycle process with alternate isothermal and adiabatic curves. In $P-V$ diagram $AB, CD, EF$ are isothermal curves at the absolute temperature $T_1, T_2$ and $T_3$ respectively and $BC, DE$ and $FA$ are adiabatic curves respectively. If $\frac{{{V_B}}}{{{V_A}}} = 2,\,\frac{{{V_D}}}{{{V_C}}} = 2$ then for cycle is shown in figure four statements are being made given below. (Figure is not drawn on scale)
Statement $1$ : Ratio of volumes $\frac{{{V_E}}}{{{V_F}}} = 4$
Statement $2$ : Magnitude of work done in isothermal compression $EF$ is $2RT_3\ ln\ (2)$
Statement $3$ : Ratio of heat supplied to gas in the process $AB$ to heat rejected by gas in process $EF$ is $\frac{{{T_1}}}{{2{T_3}}}$
Statement $4$ : Net work done by gas in the cycle $ABCDEFA$ is $(T_1 + T_2 - 2T_3) R\ ln\ (2)$
Find the number of correct statement $(s)$ given for the cyclic process followed by gas
Advanced
Download our app for free and get started
$\frac{V_{B}}{V_{A}}=2, \frac{V_{D}}{V_{C}}=2$
from a diabatic process equation $\frac{V_{E}}{V_{F}}=\frac{V_{B}}{V_{A}} \times \frac{V_{D}}{V_{C}}=2 \times 2=4$
magnitude of work done in $EF$ is
$\left|\mathrm{RT}_{3} ln \left(\frac{\mathrm{V}_{\mathrm{F}}}{\mathrm{V}_{\mathrm{E}}}\right)\right|=2 \mathrm{RT}_{3} l(2) \cdot \quad$ Ratio $\quad$ of $\quad$ heat
supplied and rejected is equal to ratio of their work done.
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1Let $\eta_{1}$ is the efficiency of an engine at $T _{1}=447^{\circ}\,C$ and $T _{2}=147^{\circ}\,C$ while $\eta_{2}$ is the efficiency at $T _{1}=947^{\circ}\,C$ and $T _{2}=47^{\circ}\,C$. The ratio $\frac{\eta_{1}}{\eta_{2}}$ will be.View Solution
- 2During an adiabatic expansion of $2\, moles$ of a gas, the change in internal energy was found $-50J.$ The work done during the process is ...... $J$View Solution
- 3View SolutionThe specific heat of a gas in an isothermal process is
- 4Consider $1 \,kg$ of liquid water undergoing change in phase to water vapour at $100^{\circ} C$. At $100^{\circ} C$, the vapour pressure is $1.01 \times 10^5 \,N - m ^2$ and the latent heat of vaporization is $22.6 \times 10^5 \,Jkg ^{-1}$. The density of liquid water is $10^3 \,kg m ^{-3}$ and that of vapour is $\frac{1}{1.8} \,kg m ^{-3}$. The change in internal energy in this phase change is nearly ............ $\,J kg ^{-1}$View Solution
- 5View SolutionWhen you make ice cubes, the entropy of water
- 6$Assertion :$ Adiabatic expansion is always accompanied by fall in temperature.View Solution
$Reason :$ In adiabatic process, volume is inversely proportional to temperature. - 7One mole of a monatomic ideal gas is taken along two cyclic processes $E \rightarrow F \rightarrow G \rightarrow E$ and $E \rightarrow F \rightarrow H \rightarrow$ E as shown in the $PV$ diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic. $Image$View Solution
Match the paths in List $I$ with the magnitudes of the work done in List $II$ and select the correct answer using the codes given below the lists.
List $I$ List $I$ $P.$ $\quad G \rightarrow E$ $1.$ $\quad 160 P_0 V_0 \ln 2$ $Q.$ $\quad G \rightarrow H$ $2.$ $\quad 36 P _0 V _0$ $R.$ $\quad F \rightarrow H$ $3.$ $\quad 24 P _0 V _0$ $S.$ $\quad F \rightarrow G$ $4.$ $\quad 31 P_0 V_0$ Codes: $ \quad \quad P \quad Q \quad R \quad S $
- 8Initial pressure and volume of a gas are $P$ and $V$ respectively. First it is expanded isothermally to volume $4V$ and then compressed adiabatically to volume $V$ . The final pressure of gas will be (given $\gamma = 3/2$ )View Solution
- 9A Carnot engine whose heat $\operatorname{sinks}$ at $27\,^{\circ} C$, has an efficiency of $25 \%$. By how many degrees should the temperature of the source be changed to increase the efficiency by $100 \%$ of the original efficiency $?$View Solution
- 10In a carnot engine, the temperature of reservoir is $527^{\circ} C$ and that of $\operatorname{sink}$ is $200 \; K$. If the workdone by the engine when it transfers heat from reservoir to sink is $12000 \; kJ$, the quantity of heat absorbed by the engine from reservoir is $\times 10^{6} \; J$View Solution