A fixed amount of a gas undergoes a thermodyn | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For a cyclic process $ABCA$,the change in internal energy is $\Delta U_{ABCA} = 0$.From the first law of thermodynamics,$Q_{ABCA} = W_{ABCA}$.This

Vedclass · Accepted Answer

can only be adiabatic

Vedclass · Answer

(B) For a cyclic process $ABCA$,the change in internal energy is $\Delta U_{ABCA} = 0$.From the first law of thermodynamics,$Q_{ABCA} = W_{ABCA}$.This can be written as: $Q_{A 	o B} + Q_{B 	o C} + Q_{C 	o A} = W_{A 	o B} + W_{B 	o C} + W_{C 	o A} \quad (1)$.Given that the heat interaction along path $B 	o C 	o A$ is equal to the work done along path $A 	o B 	o C$,we have:$Q_{B 	o C} + Q_{C 	o A} = W_{A 	o B} + W_{B 	o C} \quad (2)$.Subtracting equation $(2)$ from equation $(1)$:$Q_{A 	o B} = W_{C 	o A}$.From the $P-V$ diagram,the process $C 	o A$ is an isochoric process (constant volume),so the work done $W_{C 	o A} = 0$.Therefore,$Q_{A 	o B} = 0$.$A$ process in which the heat interaction is zero is an adiabatic process. Thus,the process $A 	o B$ is adiabatic.

$A$ fixed amount of a gas undergoes a thermodynamic process as shown in the figure,such that the heat interaction along path $B \to C \to A$ is equal to the work done by the gas along path $A \to B \to C$. Then the process $A \to B$ is:

Similar Questions

The bulk modulus of a gas is defined as $B = -V (dp / dV)$. For an adiabatic process,the variation of $B$ is proportional to $p^n$. For an ideal gas,$n$ is:

An ideal gas is found to obey $p V^{3/2} = \text{constant}$ during an adiabatic process. If such a gas initially at a temperature $T$ is adiabatically compressed to half of its initial volume, then its final temperature is

The work done in an adiabatic change in an ideal gas depends upon only

$Assertion:$ In adiabatic compression,the internal energy and temperature of the system decrease.
$Reason:$ Adiabatic compression is a slow process.

$A$ rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for the process is $TV^x =$ constant,then $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module