Given the relation V = K left( frac{P}{T} rig | vedclass

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

Question

(C) From the ideal gas law,we know that $PV = nRT$,which implies $\frac{P}{T} = \frac{nR}{V}$.Substituting this into the given equation $V = K \left(

Vedclass · Accepted Answer

isochoric process

Vedclass · Answer

(C) From the ideal gas law,we know that $PV = nRT$,which implies $\frac{P}{T} = \frac{nR}{V}$.Substituting this into the given equation $V = K \left( \frac{P}{T} ight)^{0.33}$,we get:$V = K \left( \frac{nR}{V} ight)^{0.33}$$V = K (nR)^{0.33} \cdot V^{-0.33}$Multiplying both sides by $V^{0.33}$,we get:$V^{1.33} = K' 	ext{ (where } K' = K(nR)^{0.33} 	ext{ is a constant)}$Since $V^{1.33}$ is constant,it implies that $V$ must be a constant.$A$ process in which the volume remains constant is known as an isochoric process.

Given the relation $V = K \left( \frac{P}{T} \right)^{0.33}$,where $K$ is a constant. Identify the type of thermodynamic process.

Similar Questions

$A$ diatomic gas $(\gamma = 1.4)$ does $100 \ J$ of work when it is expanded isobarically. Then the heat given to the gas is . . . . . . $J$.

$A$ diatomic gas $(\gamma = 1.4)$ does $200 \,J$ of work when it is expanded isobarically. The heat given to the gas in the process is: (in $\,J$)

An ideal gas having pressure $P$, volume $V$, and temperature $T$ undergoes a thermodynamic process in which $dW = 0$ and $dQ < 0$. Then, for the gas

$A$ diatomic gas with rigid molecules does $10 \ J$ of work when expanded at constant pressure. What would be the heat energy absorbed by the gas,in this process ..... $J$.

In a reversible isochoric change,

Vedclass Test Series

Exam Paper Generator

Online Exam Module