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

$\Delta U + \Delta W = 0$ is valid for

In an adiabatic process,the density of a diatomic gas becomes $32$ times its initial value. The final pressure of the gas is found to be $n$ times the initial pressure. The value of $n$ is

If during an adiabatic process the pressure of a mixture of gases is found to be proportional to the square of its absolute temperature,the ratio of $C_p / C_v$ for the mixture of gases is .........

$A$ thermodynamic system is compressed adiabatically, then its temperature

During a thermodynamic process, the increase in internal energy of a system is equal to the work done on the system. Which process does the system undergo?

Vedclass Test Series

Exam Paper Generator

Online Exam Module