A gas goes from state A to state B via three | vedclass

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

Question

(A) According to the first law of thermodynamics, $Q = \Delta U + W$.Since all three paths start at state $A$ and end at state $B$, the change in inte

Vedclass · Accepted Answer

$Q_1 < Q_2 < Q_3$

Vedclass · Answer

(A) According to the first law of thermodynamics, $Q = \Delta U + W$.Since all three paths start at state $A$ and end at state $B$, the change in internal energy $\Delta U$ is the same for all three paths.Therefore, the heat absorbed $Q$ depends only on the work done $W$ by the gas.The work done $W$ by the gas is equal to the area under the $P-V$ curve.From the given diagram, the area under path $1$ is the smallest, the area under path $2$ is intermediate, and the area under path $3$ is the largest.Thus, $W_1 Since $Q = \Delta U + W$ and $\Delta U$ is constant, we have $Q_1 < Q_2 < Q_3$.

$A$ gas goes from state $A$ to state $B$ via three different paths as shown in the $P-V$ diagram. If $Q_1, Q_2,$ and $Q_3$ are the heat absorbed by the gas along the three paths, then:

Similar Questions

An ideal gas follows the path shown in the figure. The net work done in the whole cycle is

The work done by a gas as it is taken in a cyclic process (shown in the graph) is (in $PV$)

In the cyclic process shown in the figure,the work done by the gas in one cycle is (in $,{P_1}{V_1}$)

In the $P-V$ diagram,$I$ is the initial state and $F$ is the final state. The gas goes from $I$ to $F$ by $(i) IAF, (ii) IBF, (iii) ICF$. The heat absorbed by the gas is

Vedclass Test Series

Exam Paper Generator

Online Exam Module