A gas can be taken from state A to state B vi | vedclass

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

Question

(D) According to the first law of thermodynamics, $\Delta U = q + w$.For path $ACB$:Heat flow into the system, $q_{ACB} = 60 	ext{ J}$.Work done by t

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(D) According to the first law of thermodynamics, $\Delta U = q + w$.For path $ACB$:Heat flow into the system, $q_{ACB} = 60 	ext{ J}$.Work done by the system, $w_{ACB} = -30 	ext{ J}$ (since work is done by the system).Change in internal energy, $\Delta U = q_{ACB} + w_{ACB} = 60 	ext{ J} - 30 	ext{ J} = 30 	ext{ J}$.Since internal energy $\Delta U$ is a state function, it remains the same for any path between $A$ and $B$.Therefore, for path $ADB$, $\Delta U = 30 	ext{ J}$.Given for path $ADB$, work done by the system, $w_{ADB} = -10 	ext{ J}$.Using $\Delta U = q_{ADB} + w_{ADB}$:$30 	ext{ J} = q_{ADB} - 10 	ext{ J}$.$q_{ADB} = 30 	ext{ J} + 10 	ext{ J} = 40 	ext{ J}$.

Similar Questions

An ideal gas expands adiabatically against vacuum. Which of the following is correct for the given process?

Calculate the internal energy change of one mole of an ideal gas during an adiabatic free expansion process from a volume of $2 \ L$ to $5 \ L$.

Assertion : For an isothermal reversible process $Q = -W$,i.e.,work done by the system equals the heat absorbed by the system.
Reason : Enthalpy change $(\Delta H)$ is zero for an isothermal process.

Which of the following statements regarding the first law of thermodynamics is correct?

When $1 \ mol$ of ice melts at $0 \ ^oC$ and a constant pressure of $1 \ atm$,the system absorbs $1440 \ cal$ of heat. If the molar volumes of ice and water are $0.0196 \ L$ and $0.0180 \ L$ respectively,calculate the change in internal energy $(\Delta U)$ of the system in $cal$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module