Work done on heating one mole of monoatomic g | vedclass

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

Question

(B) The work done in an adiabatic process is given by the formula: $\Delta W = \frac{\mu R \Delta T}{\gamma - 1}$.For $1$ mole of monoatomic gas $(\mu

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) The work done in an adiabatic process is given by the formula: $\Delta W = \frac{\mu R \Delta T}{\gamma - 1}$.For $1$ mole of monoatomic gas $(\mu = 1)$: $\Delta T = 20^{\circ} C$, $\gamma = 5/3$. Therefore, $W = \frac{1 	imes R 	imes 20}{(5/3 - 1)} = \frac{20R}{2/3} = 30R$ ... $(i)$.For $6$ moles of rigid diatomic gas $(\mu = 6)$: $\Delta T = 20^{\circ} C$, $\gamma = 7/5$. The work done $W'$ is given by: $W' = \frac{6 	imes R 	imes 20}{(7/5 - 1)} = \frac{120R}{2/5} = \frac{120R 	imes 5}{2} = 300R$.Comparing $W'$ with $W$: $W' = 300R = 10 	imes (30R) = 10W$.

Work done on heating one mole of monoatomic gas adiabatically through $20^{\circ} C$ is $W$. Then, the work done on heating $6$ moles of rigid diatomic gas through the same change in temperature is: (in $W$)

Similar Questions

For an adiabatic expansion of an ideal gas,the fractional change in its pressure is equal to (where $\gamma$ is the ratio of specific heats):

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

$A$ gas at $NTP$ is suddenly compressed to one-fourth of its original volume. If $\gamma$ is supposed to be $\frac{3}{2}$,then the final pressure is........ atmosphere.

$200 \ cc$ of an ideal gas $(\gamma = 1.5)$ expands adiabatically. If the rms speed of the gas molecules becomes half of the initial value,the final volume of the gas is (in $cc$)

The internal energy of the gas increases in:

Vedclass Test Series

Exam Paper Generator

Online Exam Module