91Ó°ÊÓ

A liter of air at atmospheric pressure is compressed adiabatically to a pressure of 7 atm.

Air is mostly diatomic.

Expression for the adiabatic process is

$P{V}^{\mathrm{γ}}=\text{constant}$

Here $P$is the pressure of air, $V$is the volume of the air, and $\mathrm{γ}$is the adiabatic constant.

In the initial and final case equation (1) can be written as

$P_i{V_i}^{\mathrm{γ}}=P_f{V_f}^{\mathrm{γ}}$

Equation (2) can be simplified as

${V_f}^{\mathrm{γ}}=\frac{P_f{V_f}^{\mathrm{γ}}}{P_f}$

Multiply $\frac{1}{\mathrm{γ}}$on both sides of equation (3)

$V_f={\left(\frac{P_i}{P_f}\right)}^{\frac{1}{\mathrm{γ}}}{V}_i$

$\begin{array}{l}{V}_f={\left(\frac{1}{7}\right)}^{\frac{5}{7}}×1\\ V_f=0.249\text{ L=2.49}×{\text{10}}^{\text{-4}}{\text{″¾}}^{\text{3}}\\ \boxed{V_f=\text{2.49}×{\text{10}}^{\text{-4}}{\text{″¾}}^{\text{3}}}\end{array}$

Hence, the required final volume is $\text{2.49}×{\text{10}}^{\text{-4}}{\text{″¾}}^{\text{3}}$.

1 liter of air is compressed adiabatically to a pressure of 7 atm. The initial volume of air is$1×{10}^{−3}{\text{″¾}}^{\text{3}}$ and the final volume is $2.49×{10}^{−4}{\text{″¾}}^{\text{3}}$.

Expression for work done on the system is

$W=−\underset{v_i}{\overset{v_f}{∫}}PdV$ …… (1)

Here $P$is the pressure, $V_i\text{ a²Ô»å }{V}_f$are initial volume, and final volume.

Expression for the adiabatic process is

role="math" localid="1651473925139" $P{V}^{\mathrm{γ}}=\text{const}$ …… (2)

Here $P$is the pressure of air,$V$ is the volume of the air, and$\mathrm{γ}$ is the adiabatic constant.

Equation (2) can be written as

$T{V}^{\mathrm{γ}−1}=\text{const}$

$T_f={\left(\frac{1×{10}^{−3}}{2.49×{10}^{−4}}\right)}^{\frac{2}{5}}×300$

$P=c{V}^{−\mathrm{γ}}$ …… (3)

Substitute $c{V}^{−\mathrm{γ}}$for $P$in equation (1)

$W=−c\underset{1×{10}^{−3}}{\overset{2.49×{10}^{−4}}{∫}}{V}^{−\mathrm{γ}}dV$ …… (4)

Solving equation (4) for$W$

$W=−c{\left[\frac{V^{−\mathrm{γ}+1}}{−\mathrm{γ}+1}\right]}_{1×{10}^{−3}}^{2.49×{10}^{−4}}$ …… (5)

Substitute $\frac{7}{5}$for $\mathrm{γ}$in equation (5)

$W=\mathrm{29.48.}c$ …… (6)

At $V=1\text{ l¾±³Ù±ð°ù}$=$1×{10}^{−3}{\text{″¾}}^{\text{3}}$and role="math" localid="1651475635356" $101325\text{ P²¹}$for $P$in equation (2)

$\begin{array}{l}P{V}^{\mathrm{γ}}=c\\ 1.013×{10}^5×{\left(1×{10}^{−3}\right)}^{\frac{7}{5}}=c\\ c=6.393\text{ P²¹.}\end{array}$

Substitute for in equation (6)

$\begin{array}{l}W=29.48×6.393\\ \boxed{W=188.44\text{ J}}\end{array}$

Hence, work done on the air is $\boxed{W=188.44\text{ J}}$.

The initial temperature is $T_i=300\text{‿é}$.

The initial volume is$V_i=1×{10}^{−3}{\text{″¾}}^{\text{3}}$ and the final volume is $V_f=2.49×{10}^{−4}{\text{″¾}}^{\text{3}}$.

Expression for adiabatic compression is

$V{T}^{\left(\frac{f}{2}\right)}=\text{const}$ …… (1)

Here, $V$is the volume of the air, $T$ is temperature and$f$ is the degree of freedom.

Equation (1) can be written for the initial and final cases as

$\begin{array}{l}{V}_i{T_i}^{\left(\frac{f}{2}\right)}=V_f{T_f}^{\left(\frac{f}{2}\right)}\\ {T_f}^{\left(\frac{f}{2}\right)}={\left(\frac{V_i}{V_f}\right)}^{\frac{2}{f}}{T}_i\end{array}$

…… (3)

Substitute 5 for $f$, $1×{10}^{−3}{\text{m}}^{\text{3}}$for $V_i$, $2.49×{10}^{−4}{\text{″¾}}^{\text{3}}$for $V_f$on equation (3)

$\begin{array}{l}{T}_f={\left(\frac{1×{10}^{−3}}{2.49×{10}^{−4}}\right)}^{\frac{2}{5}}×300\\ \boxed{T_f=523.17\text{‿é}}\end{array}$

Hence, the final temperature after compression is $\boxed{T_f=523.17\text{‿é}}$.