Question

N MOLES OF A MONOATOMIC GAS IS CARRIED AROUND THE REVERSIBLE RECTANGULAR CYCLE ABCDA AS SHOWN IN THE DIAGRAM. THE TEMPERATURE AT A IS To . THE THERMODYNAMIC EFFICIENCY OF THE CYCLE IS :
1. 15 %
2. 20 %
3. 25 %
4. 50 %

Answer 1

Answer on Question #40672, Physics, Molecular Physics

N MOLES OF A MONOATOMIC GAS IS CARRIED AROUND THE REVERSIBLE RECTANGULAR CYCLE ABCDA AS SHOWN IN THE DIAGRAM. THE TEMPERATURE AT A IS TO . THE THERMODYNAMIC EFFICIENCY OF THE CYCLE IS : 1. $15\%$ 2. $20\%$ 3. $25\%$ 4. $50\%$

Solution

In any cyclic process, the net work done by the system is equal to the area enclosed by the cyclic thermodynamic path in a p-V diagram. So

$$W _ {n e t} = (p _ {B} - p _ {A}) (V _ {D} - V _ {A}) = (2 p _ {0} - p _ {0}) (2 V _ {0} - V _ {0}) = p _ {0} V _ {0}.$$

Total heat absorbed by the cycle is the sum of the heat absorbed along path AB and the heat absorbed along path BC:

$$Q _ {A} = Q _ {A B} + Q _ {B C}.$$

The heat absorbed along path AB:

$$Q _ {A B} = C _ {V} N (T _ {B} - T _ {0}).$$

From the ideal gas state equation:

$$T _ {B} = \frac {2 p _ {0} \cdot V _ {0}}{N R},$$

and

$$T _ {0} = \frac {p _ {0} \cdot V _ {0}}{N R}.$$

The heat absorbed along path AB:

$$Q _ {A B} = C _ {V} N (T _ {B} - T _ {0}) = \frac {3}{2} R N \left(\frac {2 p _ {0} \cdot V _ {0}}{N R} - \frac {p _ {0} \cdot V _ {0}}{N R}\right) = \frac {3}{2} p _ {0} \cdot V _ {0}.$$

The heat absorbed along path BC:

$$Q _ {B C} = C _ {P} N (T _ {C} - T _ {B}).$$

From the ideal gas state equation:

$$T _ {C} = \frac {2 p _ {0} \cdot 2 V _ {0}}{N R}.$$

The heat absorbed along path AB:

$$Q _ {B C} = C _ {P} N (T _ {C} - T _ {B}) = \frac {5}{2} R N \left(\frac {2 p _ {0} \cdot 2 V _ {0}}{N R} - \frac {2 p _ {0} \cdot V _ {0}}{N R}\right) = 5 p _ {0} \cdot V _ {0}.$$

Total heat absorbed by the cycle

$$Q _ {A} = \frac {3}{2} p _ {0} \cdot V _ {0} + 5 p _ {0} \cdot V _ {0} = \frac {13}{2} p _ {0} \cdot V _ {0}.$$

Thermal efficiency of a cycle:

$$\eta = \frac {W _ {n e t}}{Q _ {A}} = \frac {p _ {0} \cdot V _ {0}}{\frac {13}{2} p _ {0} \cdot V _ {0}} = \frac {2}{13} \approx 15\%.$$

Answer: $1.15\%$ .