A cylinder contains 0.250 mol of carbon dioxide 1CO22 gas at a temperature of 27.0°C. The cylinder is provided with a frictionless piston, which maintains a constant pressure of 1.00 atm on the gas. The gas is heated until its temperature increases to 127.0°C. Assume that the CO2 may be treated as an ideal gas. (a) Draw a pV-diagram for this process. (b) How much work is done by the gas in this process? (c) On what is this work done? (d) What is the change in internal energy of the gas? (e) How much heat was supplied to the gas? (f) How much work would have been done if the pressure had been 0.50 atm?

Given,

A cylinder of $C{O}_2$gas with moles n = 0.250mol , initial temperature $T_1$= 27C = 300 K, and a constant pressure p = 1 atm , and$T_2=127C=400K$

To draw the pV diagram we first take the pressure which is constant and the temperature increases. So according to gas law the required PV diagram is:

Since the pressure is constant therefore the work done is given by:$W=p∆V$…..(i)

From ideal gas law we have: pV = nRT

As the volume changes so as the temperature, hence the gas law will become:

$\begin{array}{r}p\mathrm{\Delta }V=nR\mathrm{\Delta }T\\ \mathrm{\Delta }V=\frac{nR}{p}\mathrm{\Delta }T\\ \frac{nR}{p}\left(T_2-T_1\right)\end{array}$

Now substituting the value of in equation (i) we get:

localid="1668312242643" $$\begin{gathered} w=p\mathrm{\Delta }V \\ =p\times \frac{nR}{p}\left(T_2-T_1\right) \\ =nR\left(T_2-T_1\right) \\ =0.25\mathrm{mol}\times 8.134\mathrm{J}/\mathrm{mol}\bullet K(400-300) \\ =208\mathrm{J} \end{gathered}$$

The work done in part b is positive therefore the work is done by the gas on the piston.

Since the temperature changes from 300K to 400K, we know that internal energy depends on the temperature T. Therefore, the change in internal energy is expressed by:

$$\begin{gathered} \begin{array}{r}\mathrm{\Delta }U=n{C}_V\mathrm{\Delta }T\\ =n{C}_V\left(T_2-T_1\right)\end{array} \\ {\mathrm{C}}_V\text{of}{\mathrm{CO}}_2\text{gas is}28.46\mathrm{J}/\mathrm{molK} \end{gathered}$$

Therefore, the value of is:

$\begin{array}{r}\mathrm{\Delta }U=n{C}_V\left(T_2-T_1\right)\\ =0.25\mathrm{mol}\times 28.46\mathrm{J}/\mathrm{mol}\overline{\mathrm{J}K}(400\mathrm{K}-300\mathrm{K})\\ =712\mathrm{J}\end{array}$

Therefore, the change in internal energy is

According to the first law of thermodynamics we have:

$Q=∆U+W$

Substituting the values calculated above we have:

localid="1668312275613" $$\begin{gathered} Q=\mathrm{\Delta }U+W \\ =712\mathrm{J}+208\mathrm{J} \\ =920\mathrm{J} \end{gathered}$$

In the part b we calculated the work done, we have concluded that work done does not depend on the pressure. Therefore, the work will not change and will remain same.