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- Thermodynamics
NEET · Physics · Thermodynamics
One mole of a monatomic ideal gas is taken along two cyclic...
One mole of a monatomic ideal gas is taken along two cyclic processes \begin{cases} & \textList-I & & & \textList-II \\ \text{A.} & \text{G} \to \text{E} & & P \text{.} & 1 6 0 {\text{P}_{0} {\text{V}_{0} \text\ln 2 \\ \text{B.} & \text{G} \to \text{H} & & Q \text{.} & 3 6 {\text{P}_{0} {\text{V}_{0} \\ \text{C.} & \text{F} \to \text{H} & & R \text{.} & 2 4 {\text{P}_{0} {\text{V}_{0} \\ \text{D.} & \text{F} \to \text{G} & & S \text{.} & 3 1 {\text{P}_{0} {\text{V}_{0}\end{cases}
Options
- Aa-q;b-s;c-r;d-p;
- Ba-s;b-r;c-q;d-p;
- Ca-q;b-s;c-r;d-p;
- Da-q;b-r;c-s;d-p;
Correct answer
B. a-s;b-r;c-q;d-p;
Step-by-step solution
\text{F} \to \text{G} work done in isothermal process is nRT \ln \left(\dfrac{{\text{V}}_{\text{f}}}{{\text{V}}_{\text{i}}}\right) = 3 2 {\text{P}}_{0} {\text{V}}_{0} \text\ln \left(\dfrac{3 2 {\text{V}}_{0}}{{\text{V}}_{0}}\right)
= 3 2 {\text{P}}_{0} {\text{V}}_{0} \text\ln {2}^{5} = 1 0 {\text{P}}_{0} {\text{V}}_{0} \text{\ln 2}
\text\ln \text{G} ⟶ \text{E} \text{,} \Delta \text{W} = {\text{P}}_{0} \left(3 1 {\text{V}}_{0}\right) = 3 1 {\text{P}}_{0} {\text{V}}_{0}
\text\ln \text{G} ⟶ \text{H} work done is less than i.e.,
\text\ln \text{F} ⟶ \text{H} work done is