91影视

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.

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.

$W=-{鈭�}_{{鈭�}_i}^{v_f}1.013脳{10}^8{V}_{.}dV鈥�..\left(9\right)$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 :$W=-{鈭�}_{v_i}^{v_f}PdV\text{..... (1)}$

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.

$$\begin{gathered} P鈭�V鈥�...\left(2\right) \\ P=VA鈥�...\left(3\right) \end{gathered}$$

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)

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

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

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

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

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

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

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

$W=-{鈭�}_{{鈭�}_i}^{v_f}1.013脳{10}^8{V}_{.}dV鈥�..\left(9\right)$

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

$1脳{10}^{-3}鈥�\text{for}{V}_i\text{and}3脳{10}^{-3}\text{for}{V}_f\text{inequation}\left(9\right)$

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

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