In the P-V diagram shown,there are two adiaba | vedclass

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

Question

(B) For an adiabatic process,the relation between temperature and volume is $T V^{\gamma-1} = 	ext{constant}$.For the adiabatic path $BC$,the points

Vedclass · Accepted Answer

$\left(\frac{V_c}{V_d}\right)$

Vedclass · Answer

(B) For an adiabatic process,the relation between temperature and volume is $T V^{\gamma-1} = 	ext{constant}$.For the adiabatic path $BC$,the points $B$ and $C$ lie on the adiabatic curve connecting the isothermals $T_1$ and $T_2$. Thus:$T_1 V_b^{\gamma-1} = T_2 V_c^{\gamma-1}$$\Rightarrow \left(\frac{V_b}{V_c}ight)^{\gamma-1} = \frac{T_2}{T_1} \quad ---(1)$For the adiabatic path $AD$,the points $A$ and $D$ lie on the adiabatic curve connecting the isothermals $T_1$ and $T_2$. Thus:$T_1 V_a^{\gamma-1} = T_2 V_d^{\gamma-1}$$\Rightarrow \left(\frac{V_a}{V_d}ight)^{\gamma-1} = \frac{T_2}{T_1} \quad ---(2)$Equating $(1)$ and $(2)$:$\left(\frac{V_b}{V_c}ight)^{\gamma-1} = \left(\frac{V_a}{V_d}ight)^{\gamma-1}$$\Rightarrow \frac{V_b}{V_c} = \frac{V_a}{V_d}$Rearranging the terms,we get:$\frac{V_b}{V_a} = \frac{V_c}{V_d}$Therefore,option $(B)$ is correct.

In the $P-V$ diagram shown,there are two adiabatic parts of the same gas intersecting two isothermals at $T_1$ and $T_2$. The ratio $\left(\frac{V_b}{V_a}\right)$ is equal to

Similar Questions

$5.6 \ L$ of helium gas at $STP$ is adiabatically compressed to $0.7 \ L$. If the initial temperature of the gas is $T \ K$,the work done in the process is ($R$ is the universal gas constant in $SI$ units).

In a thermodynamic system,$\Delta U$ represents the increase in its internal energy and $dW$ is the work done by the system. Then,which of the following statements is correct?

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

The volume of an ideal gas $(\gamma=1.5)$ is changed adiabatically from $5 \ L$ to $4 \ L$. The ratio of initial pressure to final pressure is:

One mole of an ideal gas at an initial temperature of $T \ K$ does $6R \ J$ of work adiabatically. If the ratio of specific heats of this gas at constant pressure and at constant volume is $5/3$,the final temperature of the gas will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module