QJEEM23Phd724: JEE Main 2023 Physics - Class 11 JEEM23Phd724 Solution: of water at is converted into steam at by boiling at…

To find the change in internal energy, we need to consider both the heat added during the process and the work done during the process.

First, let's calculate the heat added ( $Q$ ) to convert 1 kg of water at 100°C into steam at 100°C using the latent heat of vaporization:

$Q = m \times L$

where $m = 1 \mathrm{~kg}$ and $L = 2257 \mathrm{~kJ/kg}$ :

$Q = 1 \mathrm{~kg} \times 2257 \mathrm{~kJ/kg} = 2257 \mathrm{~kJ}$

Next, let's calculate the work done ( $W$ ) on the system during the process. The work done is given by:

$W = -P \Delta V$

where $P$ is the atmospheric pressure and $\Delta V$ is the change in volume. We are given that the atmospheric pressure is $P = 1 \times 10^5 \mathrm{~Pa}$ , and the change in volume is

$\Delta V = 1.671 \mathrm{~m}^3 - 1.00 \times 10^{-3} \mathrm{~m}^3 = 1.670 \mathrm{~m}^3$ .

Now, we can calculate the work done:

$W = -(1 \times 10^5 \mathrm{~Pa})(1.670 \mathrm{~m}^3) = -167000 \mathrm{~J} = -167 \mathrm{~kJ}$

Finally, we can find the change in internal energy ( $\Delta U$ ) using the first law of thermodynamics:

$\Delta U = Q + W$

Substitute the values of $Q$ and $W$ :

$\Delta U = 2257 \mathrm{~kJ} - 167 \mathrm{~kJ} = 2090 \mathrm{~kJ}$

The change in internal energy of the system during the process is +2090 kJ.