91影视

The 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 a straight line but does not pass through the 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 does not pass through the origin. The required graph between pressure and volume is shown below:

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

Initial volume is$V_i=1\text{鈥塴颈迟别谤}$

Final volume is$V_f=3\text{鈥塴颈迟别谤}$

Initial pressure is$P_i=1\text{鈥塧迟尘}$

Expression for work done on the system is

$W=鈭�\underset{v_i}{\overset{v_f}{鈭�}}PdV$ 鈥︹�� (1)

Here $P$is the pressure, $V_i\text{鈥塧苍诲鈥夆赌�}{V}_f$are initial volume, and final volume.

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

$P鈭�V$ 鈥︹�� (2)

$P=VA$ 鈥︹�� (3)

Here $A$is the proportionality constant.

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

$\frac{P_i}{V_i}=A$ 鈥︹�� (4)

And

$\frac{P_f}{V_f}=A$ 鈥︹�� (5)

From equations (4) and (5)

$\frac{P_i}{V_i}=\frac{P_f}{V_f}$

$P_f=\frac{P_i{V}_f}{V_i}$ 鈥︹�� (6)

Substitute$1脳{10}^{鈭�3}$for$V_i$and$3脳{10}^{鈭�3}$for$V_f$and$1.013脳{10}^5$for$P_i$in equation (6)

$\begin{array}{l}{P}_f=\frac{3脳{10}^{鈭�3}脳1.013脳{10}^5}{1脳{10}^{鈭�3}}\\ P_f=3.04脳{10}^5\text{鈥塒补}\end{array}$

In the graph between pressure and volume, the slope is calculated as

$A=\frac{P_f鈭�P_i}{V_f鈭�V_i}$ 鈥︹�� (7)

Substitute $1脳{10}^{鈭�3}$for$V_i$and$3脳{10}^{鈭�3}$for $V_f$,$1.013脳{10}^5$for$P_i$and$3.04脳{10}^5\text{鈥�}$for$P_f$in equation (7)

$\begin{array}{l}A=\frac{3.04脳{10}^5鈭�1.013脳{10}^5}{3脳{10}^{鈭�3}鈭�{10}^{鈭�3}}\\ A=1.013脳{10}^8{\text{鈥塒补.m}}^{\text{-3}}\end{array}$

Hence the relation between pressure and volume is

$P=1.013脳{10}^{8}V$ 鈥�.. (8)

Substitute$1.013脳{10}^{8}V$for$P$in equation (1)

$W=鈭�\underset{v_i}{\overset{v_f}{鈭�}}1.013脳{10}^{8}V.dV$ 鈥︹�� (9)

Again substitute the upper and lower limit of the above integral i.e. $1脳{10}^{鈭�3}$for $V_i$and $3脳{10}^{鈭�3}$for $V_f$in equation (9)

$\begin{array}{l}W=鈭�\underset{{10}^{鈭�3}}{\overset{3脳{10}^{鈭�3}}{鈭�}}1.013脳{10}^{8}V.dV\\ W=鈭�1.013脳{10}^8\underset{{10}^{鈭�3}}{\overset{3脳{10}^{鈭�3}}{鈭�}}V.dV\\ W=鈭�1.013脳{10}^8{\left[\frac{V^2}{2}\right]}_{{10}^{鈭�3}}^{3脳{10}^{鈭�3}}\\ \boxed{W=鈭�405.2\text{鈥塉}}\end{array}$

Hence, work done on the gas is $\boxed{W=鈭�405.2\text{鈥塉}}$.

Initial volume and final volume are$1脳{10}^{鈭�3}$and$3脳{10}^{鈭�3}$respectively and initial and final pressure are$1.013脳{10}^5$and$3.04脳{10}^5\text{鈥�}$respectively.

Thermal energy can be expressed as

$U_{thermal}=Nf\left(\frac{1}{2}kT\right)$ 鈥︹�� (10)

Here, $f$stands for the degree of freedom, $N$stands for a number of molecules, $k$stands for Boltzmann constant, and $T$stands for temperature in Kelvin.

The equation of ideal gas is

$PV=NkT$ 鈥︹�� (11)

Here, $P$is the pressure of the gas, $V$is the volume of gas contained, $k$is Boltzmann constant, $T$is the temperature in Kelvin and $N$is the number of molecules.

Equation (10) can be simplified as by substitute$PV$for$NkT$

$U_{thermal}=\frac{f}{2}PV$ 鈥︹�� (12)

Change in thermal energy due to expansion is

$螖U=P_f{V}_f鈭�P_i{V}_i$ 鈥︹�� (13)

Equation (12) can be written as a substitute$P_f{V}_f鈭�P_i{V}_i$for $螖U$

$螖U=\frac{3}{2}螖PV=\frac{3}{2}.\left(P_f{V}_f鈭�P_i{V}_i\right)$ 鈥︹�� (14)

Substitute $1脳{10}^{鈭�3}$for $V_i$and $3脳{10}^{鈭�3}$for $V_f$,$1.013脳{10}^5$for$P_i$and$3.04脳{10}^5\text{鈥�}$for$P_f$in equation (14)

$\begin{array}{l}螖U=\frac{3}{2}\left(3.04脳{10}^5脳3脳{10}^{鈭�3}鈭�1.013脳{10}^5脳{10}^{鈭�3}\right)\\ \boxed{螖U=1216.05\text{鈥塉}}\end{array}$

Hence, a change in helium鈥檚 energy is $\boxed{螖U=1216.05\text{鈥塉}}$.

Change in energy is$螖U=1216.05\text{鈥塉}$and work done is $W=鈭�405.2\text{鈥塉}$.

The equation for the first law of thermodynamics is
$Q=螖U鈭�W$ 鈥︹�� (15)

Where $Q$is the amount of heat added or removed, $螖U$is a change in internal energy, $W$and is external work done

Substitute$1216.05\text{鈥塉}$for$螖U$and$鈭�405.2\text{鈥塉}$for$W$in equation (15)

$Q=\left(1216.05+405.2\right)\text{鈥塉}$

$\boxed{Q=1621.25\text{鈥塉}}$

Since the sign of$Q$is positive so heat is added to the system.

Hence, the amount of heat added is $\boxed{Q=1621.25\text{鈥塉}}$.

The motion of a body is related to the change in kinetic energy of the body. The kinetic energy of the body is directly proportional to the change in temperature of the body.

Kinetic energy is directly proportional to the temperature. Similarly, pressure is also directly proportional to kinetic energy. So, it can be concluded that pressure is also proportional to temperature. Basically, to increase pressure, one has to increase temperature.

Hence, to increase pressure, one has to increase the temperature.