The pressure and density of a diatomic gas wi | 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 $PV^{\gamma} = 	ext{constant}$.Since density $d = \frac{m}{V},$ we h

Vedclass · Accepted Answer

$128$

Vedclass · Answer

(C) For an adiabatic process,the relation between pressure $P$ and volume $V$ is $PV^{\gamma} = 	ext{constant}$.Since density $d = \frac{m}{V},$ we have $V = \frac{m}{d},$ which implies $V \propto \frac{1}{d}.$Substituting this into the adiabatic equation: $P \left(\frac{1}{d}ight)^{\gamma} = 	ext{constant},$ or $P \propto d^{\gamma}.$Therefore,$\frac{P'}{P} = \left(\frac{d'}{d}ight)^{\gamma}.$Given $\gamma = 7/5$ and $\frac{d'}{d} = 32,$$\frac{P'}{P} = (32)^{7/5} = (2^5)^{7/5} = 2^{5 	imes (7/5)} = 2^7.$Calculating the value,$2^7 = 128.$Thus,$\frac{P'}{P} = 128.$

The pressure and density of a diatomic gas with $\gamma = 7/5$ change adiabatically from $(P, d)$ to $(P', d').$ If $\frac{d'}{d} = 32,$ then $\frac{P'}{P}$ is

Similar Questions

If the given graph shows the logarithmic values of pressure $(P)$ and volume $(V)$ of an ideal gas, then the ratio of the specific heat capacities of the gas is

$A$ gas at $NTP$ is suddenly compressed to one-fourth of its original volume. If $\gamma$ is supposed to be $\frac{3}{2}$,then the final pressure is........ atmosphere.

In the following $P-V$ diagram, two adiabatics cut two isothermals at temperatures $T_1$ and $T_2$. The value of $\frac{V_a}{V_d}$ will be

An ideal gas undergoes a process $A \rightarrow B \rightarrow C \rightarrow A$ cycle. The process $A \rightarrow B$ is adiabatic. Calculate the work done in the process $A \rightarrow B$.

In an adiabatic expansion,which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module