The volume of an ideal gas increases 8 times | vedclass

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

Question

(D) For a reversible adiabatic process, the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.Let the init

Vedclass · Accepted Answer

$ He $

Vedclass · Answer

(D) For a reversible adiabatic process, the relation between temperature $T$ and volume $V$ is given by $TV^{\gamma-1} = 	ext{constant}$.Let the initial state be $(T_1, V_1)$ and the final state be $(T_2, V_2)$.Given: $V_2 = 8V_1$ and $T_2 = T_1/4$.Using the adiabatic relation: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Substituting the values: $T_1 V_1^{\gamma-1} = (T_1/4) (8V_1)^{\gamma-1}$.Dividing both sides by $T_1 V_1^{\gamma-1}$: $1 = (1/4) \cdot 8^{\gamma-1}$.$4 = 8^{\gamma-1}$.Expressing in base $2$: $2^2 = (2^3)^{\gamma-1} = 2^{3\gamma-3}$.Equating the exponents: $2 = 3\gamma - 3$.$3\gamma = 5$, which gives $\gamma = 5/3$.A gas with adiabatic index $\gamma = 5/3$ is a monoatomic gas.Among the given options, Helium (He) is a monoatomic gas. Therefore, the correct option is $D$.

The volume of an ideal gas increases $8$ times and the temperature becomes $1/4$ of the initial temperature during a reversible adiabatic change. If there is no exchange of heat in this process $( \Delta Q=0 )$, identify the gas from the following options:

Similar Questions

$A$ sample of gas at temperature $T$ is adiabatically expanded to double its volume. The work done by the gas in the process is (given,$\gamma = 3/2$):

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

If $\gamma = 2.5$ and the final volume is $\frac{1}{8}$ times the initial volume, then the final pressure $P'$ is equal to (Initial pressure $= P$).

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

$P-V$ plots for two gases during adiabatic processes are shown in the figure. Plots $1$ and $2$ should correspond respectively to

Vedclass Test Series

Exam Paper Generator

Online Exam Module