The amount of work done in an adiabatic expan | vedclass

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

Question

(B) For an adiabatic process,the work done $W$ by an ideal gas is given by the first law of thermodynamics: $Q = \Delta U + W$. Since the process is a

Vedclass · Accepted Answer

$\frac{R}{{\gamma - 1}}(T - {T_1})$

Vedclass · Answer

(B) For an adiabatic process,the work done $W$ by an ideal gas is given by the first law of thermodynamics: $Q = \Delta U + W$. Since the process is adiabatic,$Q = 0$,so $W = -\Delta U$.For an ideal gas,the change in internal energy is $\Delta U = nC_v \Delta T$. For $1 	ext{ mole}$ of gas,$n = 1$ and $C_v = \frac{R}{\gamma - 1}$.Thus,$W = -[C_v(T_1 - T)] = -[\frac{R}{\gamma - 1}(T_1 - T)]$.Simplifying this,we get $W = \frac{R}{\gamma - 1}(T - T_1)$.

The amount of work done in an adiabatic expansion from temperature $T$ to ${T_1}$ is

Similar Questions

Assertion $(A)$: When an ideal gas is compressed adiabatically,its temperature and the average kinetic energy of the gas molecules increase.
Reason $(R)$: The kinetic energy increases because of collisions of molecules with the moving parts of the wall.

At $N.T.P.$,one mole of a diatomic gas is compressed adiabatically to half of its volume. Given $\gamma = 1.41$,the work done on the gas will be approximately ....... $J$.

An ideal gas expands adiabatically, $(\gamma = 1.5)$. To reduce the root-mean-square (r.m.s.) velocity of the molecules $4$ times, the gas has to be expanded by a factor of: (in $times$)

The relation between volume $(V)$ and absolute temperature $(T)$ of a gas in an adiabatic process is

An adiabatic process occurs at constant

Vedclass Test Series

Exam Paper Generator

Online Exam Module