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

Two different adiabatic paths for the same gas intersect two isothermal curves as shown in the $P-V$ diagram. The relation between the ratio $\frac{V_a}{V_d}$ and the ratio $\frac{V_b}{V_c}$ is:

$A$ gas $(\gamma = 1.5)$ is suddenly compressed to $(1/4)^{th}$ of its initial volume. Find the ratio of its final pressure to its initial pressure.

In which thermodynamic process is there no exchange of heat between the system and the surroundings?

During an adiabatic process, the pressure of a gas is proportional to the cube of its temperature. The value of $C_p / C_V$ for that gas is

Vedclass Test Series

Exam Paper Generator

Online Exam Module