In an adiabatic change,the pressure P and tem | vedclass

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

Question

(C) For an adiabatic process,the relation between pressure $P$ and volume $V$ is $PV^\gamma = 	ext{constant}$.From the ideal gas equation,$PV = \mu R

Vedclass · Accepted Answer

$7/2$

Vedclass · Answer

(C) For an adiabatic process,the relation between pressure $P$ and volume $V$ is $PV^\gamma = 	ext{constant}$.From the ideal gas equation,$PV = \mu RT$,we have $V = \frac{\mu RT}{P}$.Substituting this into the adiabatic equation: $P \left( \frac{\mu RT}{P} ight)^\gamma = 	ext{constant}$.This simplifies to $P^{1-\gamma} T^\gamma = 	ext{constant}$,or $P \propto T^{\frac{\gamma}{\gamma-1}}$.For a diatomic gas,the adiabatic exponent $\gamma = 7/5 = 1.4$.Therefore,$C = \frac{\gamma}{\gamma-1} = \frac{7/5}{7/5 - 1} = \frac{7/5}{2/5} = 7/2$.

In an adiabatic change,the pressure $P$ and temperature $T$ of a diatomic gas are related by the relation $P \propto T^C$,where $C$ is equal to:

Similar Questions

Heat is not being exchanged in a body. If its internal energy is increased,then

$A$ cycle tyre bursts suddenly. This process represents an:

The $P-V$ diagram of a diatomic ideal gas system undergoing a cyclic process is shown in the figure. The work done during the adiabatic process $CD$ is (use $\gamma=1.4$) (in $J$):

$A$ given system undergoes a change in which the work done by the system equals the decrease in its internal energy. The system must have undergone an

$A$ diatomic gas undergoes adiabatic change. Its pressure $P$ and temperature $T$ are related as $P \propto T^{x}$ where the value of $x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module