The mean kinetic energy of monoatomic gas mol | 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 $T$ and volume $V$ is given by $T V^{\gamma-1} = 	ext{constant}$.For a monoatomic gas,t

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) For an adiabatic process,the relation between temperature $T$ and volume $V$ is given by $T V^{\gamma-1} = 	ext{constant}$.For a monoatomic gas,the adiabatic index $\gamma = \frac{5}{3}$.Given that the gas is compressed to $1/8$ of its initial volume,we have $V_1 = V$ and $V_2 = \frac{V}{8}$.Using the adiabatic relation: $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Substituting the values: $T_1 V^{\frac{5}{3}-1} = T_2 \left(\frac{V}{8}ight)^{\frac{5}{3}-1}$.$T_1 V^{\frac{2}{3}} = T_2 \left(\frac{V}{8}ight)^{\frac{2}{3}}$.$\frac{T_2}{T_1} = \left(\frac{V}{V/8}ight)^{\frac{2}{3}} = 8^{\frac{2}{3}} = (2^3)^{\frac{2}{3}} = 2^2 = 4$.The mean kinetic energy $\langle E angle$ of a gas molecule is given by $\langle E angle = \frac{3}{2} k_B T$,which implies $\langle E angle \propto T$.Therefore,$\frac{\langle E_2 angle}{\langle E_1 angle} = \frac{T_2}{T_1} = 4$.

Similar Questions

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

In a thermodynamic system,$\Delta U$ represents the increase in internal energy and $W$ represents the work done by the system. Which of the following statements is true?

An ideal gas is expanded adiabatically at an initial temperature of $300 \ K$ so that its volume is doubled. The final temperature of the hydrogen gas is $(\gamma = 1.40)$.

For an adiabatic compression of $1 \text{ kmol}$ of gas,$146 \text{ kJ}$ of work is done. During this process,the temperature of the gas increases by $7 \text{ °C}$. The gas is ........

$5$ moles of Hydrogen $\left(\gamma=\frac{7}{5}\right)$ initially at $S.T.P.$ are compressed adiabatically so that its temperature becomes $400^{\circ} C$. The increase in the internal energy of the gas in kilo-joules is $\left(R=8.30 \ J \ mol^{-1} \ K^{-1}\right)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module