Question

A cylinder contains 0.0100 mol of helium at T = 27.0°C. (a) How much heat is needed to raise the temperature to 67.0°C while keeping the volume constant? Draw a pV-diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from 27.0°C to 67.0°C? Draw a pV-diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare?

Short answer

(a) 4.99 J of heat is required to raise the temperature to 67.0 C . And the required PV diagram of the process is:

(b) If instead the pressure of the helium is kept constant the heat required is 8.31J to raise the temperature from to $27.0°C$to $67.0°C$, and the required PV diagram is:

(c) In Part the additional heat goes to the work done as the pressure is kept constant.

(d) If the gas is ideal, change in its internal energy in both part a and b remains the same that is 4.99J

Step-by-step solution

 To Calculate the heat added

Given,

Number of moles n = 0.0100mol and the initial temperature$T_1=27C=300K$.

Now, at constant volume we can use the molar heat capacity at a constant volume$C_v$, and to find the heat added to rise the temperature from to $T_{1}to{T}_2=67C=340K$, we use:

$\begin{array}{r}Q=n{C}_v\mathrm{\Delta }T\\ =n{C}_v\left(T_2-T_1\right)\end{array}$

We know that the $C_v$of helium is 12.47 J/mol K, and now substituting the values of$n,C_v,T_2$and$T_1$we get:

$\begin{array}{r}Q=n{C}_v\left(T_2-T_1\right)\\ =0.0100\mathrm{mol}\times 12.47\mathrm{J}/\mathrm{molK}(340\mathrm{K}-300\mathrm{K})\\ =4.99\mathrm{J}\end{array}$

Therefore, the heat added is 4.99 J

Now, to draw the PV diagram for the process, we have the volume is constant while the temperature increases, therefore according to ideal gas law the pressure will also increase, and the resultant PV diagram is:

Required heat to increase the temperature

Here the pressure is constant, so using the molar heat capacity at constant pressure$C_p$ instead of$C_v$ in the above expression we have:

$Q=n{C}_p\left(T_2-T_1\right)$

Here, the$C_p$ for helium is 20.78 J/molK , substituting the values we get:

$\begin{array}{r}Q=n{C}_p\left(T_2-T_1\right)\\ =0.0100\mathrm{mol}\times 20.78\mathrm{J}/\mathrm{molK}(340\mathrm{K}-300\mathrm{K})\\ =8.31\mathrm{J}\end{array}$

And the required PV diagram where the pressure is constant and volume increasing is shown below:

Calculating the change of internal energy

The change in internal$∆U$can be calculated by using the first law of thermodynamics, where:

$∆U=Q-W$

However, the volume is kept constant therefore the work done is zero.

$W=0$

Now,

localid="1668312378353" $$\begin{gathered} \mathrm{\Delta }{U}_a=Q-W \\ =4.99-0 \\ =4.99\mathrm{J} \end{gathered}$$

In part b the$∆U_b$will also be 4.99 J because the increase of Q in part b in used to increase the work done and not the internal energy.

Hence,$∆U$is same for both part and b that is 4.99J