A monoatomic ideal gas,initially at temperatu | vedclass

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

Question

(D) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = 	ext{constant}$.Thus,$T_1 V_1^{\

Vedclass · Accepted Answer

$(\frac{L_2}{L_1})^{2/3}$

Vedclass · Answer

(D) For an adiabatic process,the relationship between temperature $T$ and volume $V$ is given by $T V^{\gamma - 1} = 	ext{constant}$.Thus,$T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1}$,which implies $\frac{T_1}{T_2} = (\frac{V_2}{V_1})^{\gamma - 1}$.Since the gas is monoatomic,the adiabatic index $\gamma = 5/3$. Therefore,$\gamma - 1 = 5/3 - 1 = 2/3$.The volume of the gas in a cylinder is $V = A 	imes L$,where $A$ is the cross-sectional area and $L$ is the length of the gas column.Substituting $V_1 = A L_1$ and $V_2 = A L_2$,we get $\frac{V_2}{V_1} = \frac{A L_2}{A L_1} = \frac{L_2}{L_1}$.Substituting these into the temperature ratio equation: $\frac{T_1}{T_2} = (\frac{L_2}{L_1})^{2/3}$.

Similar Questions

In an adiabatic change,the pressure $P$ and temperature $T$ of a diatomic gas are related by the relation $P \propto T^C$,where $C$ is equal to:

The pressure and density of a given mass of a diatomic gas $\left(\gamma = \frac{7}{5}\right)$ change adiabatically from $(P, d)$ to $(P^{\prime}, d^{\prime})$. If $\frac{d^{\prime}}{d} = 32$, then $\frac{P^{\prime}}{P}$ is $(\gamma = \text{ratio of specific heats})$.

$A$ rigid diatomic gas $(\gamma = 7/5)$ is compressed adiabatically to volume $(V_i/32)$, where $V_i$ is the initial volume. The initial temperature of the gas is $T_i \text{ K}$ and the final temperature is $x T_i \text{ K}$. The value of $x$ is

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:

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:

Vedclass Test Series

Exam Paper Generator

Online Exam Module