Consider 1 ,kg of liquid water undergoing change in phase to water vapour at 100^{circ} C . At 100^{

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11.Thermodynamics

normal

Consider $1 \,kg$ of liquid water undergoing change in phase to water vapour at $100^{\circ} C$. At $100^{\circ} C$, the vapour pressure is $1.01 \times 10^5 \,N - m ^2$ and the latent heat of vaporization is $22.6 \times 10^5 \,Jkg ^{-1}$. The density of liquid water is $10^3 \,kg m ^{-3}$ and that of vapour is $\frac{1}{1.8} \,kg m ^{-3}$. The change in internal energy in this phase change is nearly ............ $\,J kg ^{-1}$

$1.8 \times 10^5$

$20.8 \times 10^5$

$22.6 \times 10^5$

$11.3 \times 10^5$

(KVPY-2009)

Solution

(b)

Change in volume of $1 \,kg$ of water due to phase change is

$\Delta V=V_{\text {steam }}-V_{\text {water }}$

$=\frac{m_{\text {seam }}-m_{\text {water }}}{\rho_{\text {steam }}}-\rho_{\text {water }}$

$=\frac{1}{(1 / 18)}-\frac{1}{1000}$

$=1800-0.001=1799 \,m ^3$

Work done against atmospheric pressure during phase change is

$\Delta W =p \Delta V$

$=101 \times 10^5 \times 1799$

$=181699 \,J$

Heat absorbed during phase change is

$\Delta Q =m L$

$=1 \times 22.6 \times 10^5$

$=2260000 \,J$

So, change in internal energy is

$\Delta U =\Delta Q-\Delta W$

$=2260000-181699$

$=2078301 J \,kg$

$=208 \times 10^5 \,J kg ^{-1}$

Standard 11

Physics

An ideal gas is taken through the cycle $A \to B \to C \to A$ , as shown in the figure. If the net heat supplied to the gas in the cycle is $5\ J$, the work done by the gas in the process $C \to A$ is …. $J$

medium

When a system is taken from a state $i$ to $f$ along the path $iaf$ (as shown in the figure). $Q = 50\, cal$ and $W = 20\, cal$ ; along path $ibf,$ $Q = 36\, cal.$

$(i)$ What is $W$ along path $ibf$ ?

$(ii)$ If $W = 13$ cal for path $fi$, what is $Q$ for the path $fi$ ?

$(iii)$ Take $E_{int,i} = 10\,\, cal$ then what is $E_{int,f}$ ?

hard

(AIIMS-2009)

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

$(A)$ the process during the path $\mathrm{A} \rightarrow \mathrm{B}$ is isothermal

$(B)$ heat flows out of the gas during the path $\mathrm{B} \rightarrow \mathrm{C} \rightarrow \mathrm{D}$

$(C)$ work done during the path $\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}$ is zero

$(D)$ positive work is done by the gas in the cycle $ABCDA$

normal

(IIT-2009)

For the given cyclic process $CAB$ as shown for a gas, the work done is ….. $J$

easy

(JEE MAIN-2019)

In an adiabatic process $90J$ of work is done on the gas. The change in internal energy of the gas is ……. $J$

easy

Solution

Similar Questions

An ideal gas is taken through the cycle $A \to B \to C \to A$ , as shown in the figure. If the net heat supplied to the gas in the cycle is $5\ J$, the work done by the gas in the process $C \to A$ is …. $J$

For the given cyclic process $CAB$ as shown for a gas, the work done is ….. $J$

In an adiabatic process $90J$ of work is done on the gas. The change in internal energy of the gas is ……. $J$

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