If Q , E and W denote respectively heat added, change in internal energy and work done by a closed c

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11.Thermodynamics

easy

If $Q , E$ and $W$ denote respectively the heat added, change in internal energy and the work done by a closed cycle process, then

$W = 0$

$ Q = 0$

$E= 0$

$Q=W=0$

(AIPMT-2008)

Solution

Internal energy depends only on the initial and final states of temperature and not on the path. In a cyclic process, as initial and final states are the same, change in internal energy is zero. Hence $E$ is $\Delta U$, the change in internal energy.

Standard 11

Physics

An ideal gas undergoes the process $1 \rightarrow 2$ as shown in the figure, the heat supplied and work done in the process is $\Delta \,\,Q$ and $\Delta \,\,W$ respectively. The ratio $\Delta \,\,Q :$ $\Delta \,\,W$ is

hard

A thermodynamic system is taken from state $A$ to $B$ along $ACB$ and is brought back to $A$ along $BDA$ as shown in the $PV$ diagram. The net work done during the complete cycle is given by the area

medium

(AIPMT-1992)

When a system is taken from state $i$ to a state $f$ along path $iaf, \,Q = 50\,J$ and $W = 20J.$ Along path $ibf, \,Q = 35J.$ If $W = – 13J$ for the curved return path $f i, Q$ for this path is …… $J$

medium

In pressure-volume diagram given below, the isochoric, isothermal, and isobaric parts respectively, are

medium

The figure shows the $P-V$ plot of an ideal gas taken through a cycle $ABCDA.$ The part $ABC$ is a semicircle and $CDA$ is half of an ellipse. Then,

medium

If $Q , E$ and $W$ denote respectively the heat added, change in internal energy and the work done by a closed cycle process, then

Solution

Similar Questions

An ideal gas undergoes the process $1 \rightarrow 2$ as shown in the figure, the heat supplied and work done in the process is $\Delta \,\,Q$ and $\Delta \,\,W$ respectively. The ratio $\Delta \,\,Q :$ $\Delta \,\,W$ is

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