Question:
In a Carnot's cycle, if $V_1$ and $V_2$ are respectively the initial and final volumes of the working substance during isothermal expansion process, $V_3$ and $V_4$ are respectively the initial and final volumes of the working substance during isothermal compression process, then the relation among $V_1, V_2, V_3$ and $V_4$ is
In a Carnot's cycle, if $V_1$ and $V_2$ are respectively the initial and final volumes of the working substance during isothermal expansion process, $V_3$ and $V_4$ are respectively the initial and final volumes of the working substance during isothermal compression process, then the relation among $V_1, V_2, V_3$ and $V_4$ is
Show Hint
A standard Carnot cycle result is:
\[
\frac{V_2}{V_1}
=
\frac{V_3}{V_4}
\]
which directly gives
\[
V_1V_3=V_2V_4
\]
Updated On: Jun 17, 2026
- $V_1+V_3=V_2+V_4$
- $V_1V_2=V_3V_4$
- $V_1V_3=V_2V_4$
- $V_1+V_2=V_3+V_4$
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
In a Carnot cycle, the adiabatic expansion and compression processes satisfy
\[
TV^{\gamma-1}=constant
\]
Combining the two adiabatic relations leads to a special volume relationship.
Step 1: Write adiabatic relations.
For adiabatic expansion, \[ T_HV_2^{\gamma-1} = T_LV_3^{\gamma-1} \] For adiabatic compression, \[ T_HV_1^{\gamma-1} = T_LV_4^{\gamma-1} \]
Step 2: Divide the two equations.
\[ \frac{V_2^{\gamma-1}} {V_1^{\gamma-1}} = \frac{V_3^{\gamma-1}} {V_4^{\gamma-1}} \] \[ \left(\frac{V_2}{V_1}\right) = \left(\frac{V_3}{V_4}\right) \]
Step 3: Obtain the required relation.
\[ V_1V_3 = V_2V_4 \] Hence, \[ \boxed{V_1V_3=V_2V_4} \]
Step 1: Write adiabatic relations.
For adiabatic expansion, \[ T_HV_2^{\gamma-1} = T_LV_3^{\gamma-1} \] For adiabatic compression, \[ T_HV_1^{\gamma-1} = T_LV_4^{\gamma-1} \]
Step 2: Divide the two equations.
\[ \frac{V_2^{\gamma-1}} {V_1^{\gamma-1}} = \frac{V_3^{\gamma-1}} {V_4^{\gamma-1}} \] \[ \left(\frac{V_2}{V_1}\right) = \left(\frac{V_3}{V_4}\right) \]
Step 3: Obtain the required relation.
\[ V_1V_3 = V_2V_4 \] Hence, \[ \boxed{V_1V_3=V_2V_4} \]
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