Consider the given diagram. An ideal gas is c | vedclass

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

Question

(A) This process is known as free expansion of an ideal gas.Since the system is thermally isolated, there is no heat exchange with the surroundings, s

Vedclass · Accepted Answer

$T$

Vedclass · Answer

(A) This process is known as free expansion of an ideal gas.Since the system is thermally isolated, there is no heat exchange with the surroundings, so $Q = 0$.Since the gas expands into a vacuum, there is no external pressure against which the gas does work, so the work done $W = 0$.According to the first law of thermodynamics, $\Delta U = Q - W$.Since $Q = 0$ and $W = 0$, the change in internal energy $\Delta U = 0$.For an ideal gas, the internal energy depends only on its temperature $(U \propto T)$.Since $\Delta U = 0$, the temperature of the ideal gas remains constant.Therefore, the final temperature attained at equilibrium will be $T$.

Similar Questions

Name the process which is associated with dissipated energy process and reduce the efficiency.

Which of the following is $NOT$ a reversible process?

If a Van der Waals gas expands freely,then the final temperature is:

Which thermodynamic process is quasi-static and does not dissipate energy?

Vedclass Test Series

Exam Paper Generator

Online Exam Module