Question

Imagine some helium in a cylinder with an initial volume of $1litre$and an initial pressure of $1atm.$Somehow the helium is made to expand to a final volume of $3litres$,in such a way that its pressure rises in direct proportion to its volume.

(a) Sketch a graph of pressure vs. volume for this process.

(b) Calculate the work done on the gas during this process, assuming that there are no "other" types of work being done.

(c) Calculate the change in the helium's energy content during this process.

(d) Calculate the amount of heat added to or removed from the helium during this process.

(e) Describe what you might do to cause the pressure to rise as the helium expands.

Short answer

a. The graph is :

b. Work done on the gas is,$W=-405.2\mathrm{J}.$

c. Change in helium's energy is, $\Delta U=1216.05\mathrm{J}.$

d. The amount of heat added to helium during this process is $1621.25\text{J.}$

e. To increase the pressure as helium expands we have to increase the temperature.

Step-by-step solution

Given information

Some helium in a cylinder with an initial volume of $1litre$and an initial pressure of $1atm.$Now, the helium is made to expand to a final volume of$3litres,$ in such a way that its pressure rises in direct proportion to its volume.

Calculation

a. Pressure versus volume diagram can be drawn by taking pressure on y -axis and volume on x-axis. In this process, since the pressure is directly proportional to the volume so the final pressure vs. volume graph is straight line but not passing through origin.

Since the pressure is directly proportional to the volume of helium gas so for this process, the graph between pressure and volume is a straight line but not passing through origin. The required graph between pressure and volume is shown below:

Hence, the required graph between pressure and volume is a straight line but not passing through origin.

Continuation of calculation

$W=-{\int }_{{\nabla }_i}^{v_f}1.013\times {10}^8{V}_{.}dV\dots ..\left(9\right)$b. Initial volume is $V_i=1litre$

Final volume is $V_f=3litre$

Initial pressure is $P_i=1\mathrm{atm}$

Expression for work done on the system is :$W=-{\int }_{v_i}^{v_f}PdV\text{..... (1)}$

Here P is pressure, $V_i\text{and}{V}_f$are initial volume and final volume.

Pressure is directly proportional to the volume as given i.e.

$$\begin{gathered} P\propto V\dots ...\left(2\right) \\ P=VA\dots ...\left(3\right) \end{gathered}$$

Here A is proportionality constant.

From equation (3) it can be simplified as for initial and final case

$$\begin{gathered} \frac{p_t}{V_t}=A\text{..... (4)} \\ \text{And} \\ \frac{p_f}{V_t}=A\text{...... (5)} \end{gathered}$$

From equation (4) and (5)

$$\begin{gathered} \frac{P_t}{V_d}=\frac{p_f}{V_f} \\ {P}_f=\frac{P_d{V}_f}{V_i}\dots \dots \left(6\right) \end{gathered}$$

Substitute :$1\times {10}^{-3}\text{for}{V}_i\text{and}3\times {10}^{-3}\text{for}{V}_f\text{and}1.013\times {10}^5\text{for}{P}_i\text{inequation}\left(6\right)$

In the graph between pressure and volume, slope is calculated as :$A=\frac{{\mathit{p}}_f-{\mathit{p}}_f}{V_f-V_f}\dots \dots \left(7\right)$

Substitute : $1\times {10}^{-3}\text{for}{V}_i\text{and}3\times {10}^{-3}\text{for}{V}_f,1.013\times {10}^5\text{for}{P}_l\text{and}3.04\times {10}^5\text{for}{P}_f\text{inequation}\left(7\right)$

$$\begin{gathered} A=\frac{3.04\times {10}^3-1.013\times {10}^3}{3\times {10}^{-3}-{10}^{-3}} \\ A=1.013\times {10}^8{\mathrm{Pa}}^{-.{\mathrm{m}}^{-3}} \end{gathered}$$

Hence the relation between pressure and volume is : $P=1.013\times {10}^8{V}_{\dots }\left(8\right)$

Substitute : role="math" localid="1650445807757" $1.013\times {10}^{8}V$for P in equation (1)

$W=-{\int }_{{\nabla }_i}^{v_f}1.013\times {10}^8{V}_{.}dV\dots ..\left(9\right)$

Again substitute upper and lower limit of the above integral i.e.

$1\times {10}^{-3}\hspace{0.17em}\text{for}{V}_i\text{and}3\times {10}^{-3}\text{for}{V}_f\text{inequation}\left(9\right)$

$$\begin{gathered} W=-{\int }_{{10}^{-3}}^{3\times {10}^{-3}}1.013\times {10}^{8}V.dV \\ W=-1.013\times {10}^8{\int }_{{10}^{-3}}^{3\times {10}^{-3}}V.dV \\ W=-1.013\times {10}^8{\left[\frac{V^2}{2}\right]}_{{10}^{-3}}^{3\times {10}^{-3}} \\ W=-405.2\mathrm{J} \end{gathered}$$

Hence, work done on the gas is$W=-405.2\mathrm{J}.$