Question

The process abc shown in the pV-diagram in Fig. E19.11 involves 0.0175 mol of an ideal gas. (a) What was the lowest temperature the gas reached in this process? Where did it occur? (b) How much work was done by or on the gas from a to b? From b to c? (c) If 215 J of heat was put into the gas during abc, how many of those joules went into internal energy?

Short answer

(a) The lowest temperature the gas reached is 278K and it occurred at point a

(b) Total work done from a to b is zero.

(c) If 215J of heat was put into the gas then 162J heat went into the gas.

Step-by-step solution

Lowest temperature

Given,

It is given an ideal gas with moles$n=0.0175mol$ From the graph we can see that at point the pressure$p=0.2atm$ and the volume is$V=2L$

Now, applying the ideal gas formula and solve for T we get:

$T=\frac{pV}{nR}$, where$R$ is gas constant equals to$8.314J$ and the pressure is converted to Pa that is$1.013\times {10}^{5}Pa$

$$\begin{gathered} =\frac{\left(0.2atm\times 1.013\times {10}^{5}Pa\right)\left(2\times {10}^{-3}{m}^3\right)}{0.0175mol\times 8.314J/mol\bullet K} \\ =278K \end{gathered}$$

Therefore, the lowest temperature isdata-custom-editor="chemistry" $278\mathrm{K}$

Calculating work done and heat absorbed

Since in the process from a to b the volume is constant therefore the work done is $W_{ab}=0$

Now, in the process from b to c the pressure decreases from 0.5atmto 0.3 atmand the volume increases from 2L to 6L .

Therefore, the work done from b to c is given by:

$$\begin{gathered} W_{ab}=\frac{1}{2}\left[\left(0.3+0.5\right)\left(1.013\times {10}^{5}Pa\right)\right]\times [\left(6-2\right)\times {10}^{-3}{m}^3] \\ =162J \end{gathered}$$

Therefore, the work done is$162J$

Now it is given the value to heat added$Q=215J$ . So, to calculate the heat added we use the first law of thermodynamics

$$\begin{gathered} ∆U=Q-W \\ =215J-162J \\ =53J \end{gathered}$$

Therefore, the heat absorbed is$53J$