Question

Two moles of an ideal gas are heated at constant pressure from T = 27°C to T = 107°C. (a) Draw a pV-diagram for this process. (b) Calculate the work done by the gas

Short answer

(a) The diagram for the process is:

(b) The work done by the two molecules of an ideal gas when heated at constant pressure is 1330.3J

Step-by-step solution

About Ideal Gas equation and work done in a volume change at constant pressure

The Ideal gas equation is given by:

Where, p is the pressure of the gas,V The volume of the gas,number of moles, R is Ideal gas constant or 8.314472J/mol.K , and T is temperature.

The work done by gas in a volume change at constant pressure is:

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

Where, p is pressure of the gas, $V_2$is final volume, $V_1$is initial volume.

The pV diagram of the process

Therefore, the pV diagram of the process is liner.

Work done by the two moles of an ideal gas

Given,$T_2=102°C$

$T_1=27°C$

From above we have seen that the equation of an Ideal gas is: PV = nRT , here n,P,R are constant

$$\begin{gathered} P∆V=nR∆T \\ W=p\left(V_2-V_1\right) \\ P∆V=nR∆T \\ =2\left(8.314472J/(mol.K)\right)(107-27)K \\ =1330.3J \end{gathered}$$

Therefore, the work done by the two molecules of the ideal gas is 1330.3J