Question

Figure shows two paths that may be taken by a gas from an initial point i to a final point f. Path 1 consists of an isothermal expansion (work is $\mathbf{50}\mathbf{\text{\hspace{0.17em}J}}$in magnitude), an adiabatic expansion (work is $\mathbf{40}\mathbf{\text{\hspace{0.17em}J}}$in magnitude), an isothermal compression (work is 30Jin magnitude) and then an adiabatic compression (work is 25Jin magnitude). What is the change in the internal energy of the gas if the gas goes from point i to point f along path 2?

Short answer

The change in internal energy of the gas if gas goes from point i to f along path 2 is$-15\text{J}$ .

Step-by-step solution

Given

• Work done in isothermal expansion is$50\text{\hspace{0.17em}J}$
• Work done in isothermal compression is$30\text{\hspace{0.17em}J}$
• Work done in adiabatic expansion is$40\text{\hspace{0.17em}J}$
• Work done in adiabatic compression is$25\text{\hspace{0.17em}J}$

Understanding the concept

The expression for the heat from the first law of thermodynamics is given by,

$Q=W+\Delta U$

Here Q is the heat required, W is the work done, $\Delta U$is the change in internal energy.

So, the first law of thermodynamics gives the relation between heat, work and internal energy.

Calculate the change in the internal energy of the gas if the gas goes from point i to point f along path 2

Since internal energy is a state function, change in internal energy along path 1 and 2 are equal.

$\Delta U_1=\Delta U_2$

For adiabatic expansion,

Work is done by the system, so work done is positive.

$W=40\text{J}$

First law of thermodynamics gives,

$Q=W+\Delta U_1$

For adiabatic process$\text{Q}=0$

So,

$\begin{array}{c}0=40+\Delta U_1\\ \Delta U_1=-40\text{\hspace{0.17em}J}\end{array}$

For adiabatic compression,

Work is done on the system, so work done is negative.

$W=-25\text{\hspace{0.17em}J}$

$Q=W+\Delta U_2$

For adiabatic process$Q=0$

So,

$\begin{array}{c}0=-25+\Delta U_2\\ \Delta U_2=25\text{\hspace{0.17em}J}\end{array}$

So total change in internal energy is

$\begin{array}{c}\Delta U=\Delta U_1+\Delta U_2\\ =-40+25\\ \Rightarrow \Delta U=-15\text{\hspace{0.17em}J}\end{array}$

Therefore, the change in internal energy along path 2 is$-15\text{\hspace{0.17em}J.}$