A diatomic gas undergoes adiabatic change. It | vedclass

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

Question

(A) For an adiabatic process,the relationship between pressure $P$ and temperature $T$ is given by $T^{\gamma} P^{1-\gamma} = 	ext{constant}$,which c

Vedclass · Accepted Answer

$3.5$

Vedclass · Answer

(A) For an adiabatic process,the relationship between pressure $P$ and temperature $T$ is given by $T^{\gamma} P^{1-\gamma} = 	ext{constant}$,which can be rewritten as $P \propto T^{\frac{\gamma}{\gamma-1}}$.For a diatomic gas,the adiabatic index $\gamma = \frac{7}{5} = 1.4$.Substituting the value of $\gamma$ into the exponent $x = \frac{\gamma}{\gamma-1}$:$x = \frac{1.4}{1.4-1} = \frac{1.4}{0.4} = \frac{14}{4} = 3.5$.Thus,$P \propto T^{3.5}$,so the value of $x$ is $3.5$.

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

Similar Questions

Can the temperature of an isolated system of gas be changed?

An ideal gas undergoes an adiabatic process obeying the relation $PV^{4/3} = \text{constant}$. If its initial temperature is $300 \ K$ and its pressure is increased up to four times its initial value, then the final temperature is (in Kelvin):

$A$ monoatomic gas at pressure $P$ and volume $V$ is suddenly compressed to one-eighth of its original volume. The final pressure at constant entropy will be $.....P$.

By what factor should the volume of an ideal gas $(\gamma = 1.5)$ be increased through an adiabatic expansion so that its $rms$ speed becomes half?

Compressed air in the tube of a bicycle wheel at normal temperature suddenly starts coming out from a puncture. The air inside:

Vedclass Test Series

Exam Paper Generator

Online Exam Module