Shows a pV-diagram for an ideal gas in which its absolute temperature at b is one-fourth of its absolute temperature at a. (a) What volume does this gas occupy at point b? (b) How many joules of work was done by or on the gas in this process? Was it done by or on the gas? (c) Did the internal energy of the gas increase or decrease from a to b? How do you know? (d) Did heat enter or leave the gas from a to b? How do you know?

The work done when volume changes at constant pressure is:

$$\begin{gathered} W=pa∆x=p∆V \\ W=p\left(V_2-V_1\right)=p∆V \end{gathered}$$

Where, p stands for the pressure,$V_2$ stands for final volume,$V_1$ stands for initial volume.

When gas expands it makes works on its surrounding.

Now from Ideal Gas equation we get: PV = nRT

Where P stands for pressure, V stands for volume, n stands for number of moles, T is temperature and R is the gas constants equals to 8.314472J/mol.K

From Ideal gas equation we get:

$PV=nRT$, where, n,R,P are constant.

$\frac{V_b}{V_a}=\frac{T_b}{T_a}$solving for$V_b$ we get

$$\begin{gathered} V_b=\frac{T_b}{T_a}{V}_a \\ =\frac{{\displaystyle \frac{1}{4}}{T}_a}{T_a}{V}_a \\ =\frac{1}{4}{V}_a \\ =0.25\times 0.5 \\ =0.125L \end{gathered}$$

Therefore, the volume of the gas is 0.125L

We know that$1atm=101325Pa$

$$\begin{gathered} W=p∆V=p\left(V_2-V_1\right) \\ =1.5\times \left(101325\right)\left(\frac{1}{4}\times 0.5-0.5\right)\times {10}^{-3}{m}^3 \\ =-57J \end{gathered}$$

Therefore, the work done of the gas is -57J

Since, the work done is negative, hence the internal energy decreases and the heat leaves the gas.