When unit mass of water boils to become steam | vedclass

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

Question

(B) According to the first law of thermodynamics,the change in internal energy is given by $\Delta U = Q - W$.The work done $W$ by the system during t

Vedclass · Accepted Answer

$Q + p_0 \left( \frac{1}{\rho_1} - \frac{1}{\rho_2} \right)$

Vedclass · Answer

(B) According to the first law of thermodynamics,the change in internal energy is given by $\Delta U = Q - W$.The work done $W$ by the system during the phase change at constant atmospheric pressure $p_0$ is $W = p_0 \Delta V$.For a unit mass $(m = 1)$,the volume of water is $V_1 = \frac{1}{\rho_1}$ and the volume of steam is $V_2 = \frac{1}{\rho_2}$.Thus,the work done is $W = p_0 (V_2 - V_1) = p_0 \left( \frac{1}{\rho_2} - \frac{1}{\rho_1} \right)$.Substituting this into the first law equation: $\Delta U = Q - p_0 \left( \frac{1}{\rho_2} - \frac{1}{\rho_1} \right) = Q + p_0 \left( \frac{1}{\rho_1} - \frac{1}{\rho_2} \right)$.

When unit mass of water boils to become steam at $100\,^{\circ}C$,it absorbs $Q$ amount of heat. The densities of water and steam at $100\,^{\circ}C$ are $\rho_1$ and $\rho_2$ respectively and the atmospheric pressure is $p_0$. The increase in internal energy of the water is:

Similar Questions

In a cyclic process,the work done by the system is

$A$ system is taken from state $A$ to state $B$ along two different paths. The heat absorbed and work done by the system along these two paths are $Q_1, Q_2$ and $W_1, W_2$ respectively,then

If $\Delta Q$ and $\Delta W$ represent the heat supplied to the system and the work done on the system respectively,then the first law of thermodynamics can be written as,where $\Delta U$ is the change in internal energy.

In a given process,for an ideal gas,$\Delta W = 0$ and $\Delta Q < 0$. Then,for the gas,

Calculate the change in internal energy when $1 \,g$ of water is converted into steam at atmospheric pressure $(1.013 \times 10^{5} \,Pa)$. Given the latent heat of vaporisation is $2256 \,J/g$, the volume of $1 \,g$ of water is $1 \,cm^{3}$, and the volume of $1 \,g$ of steam is $1671 \,cm^{3}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module