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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

$P{V}^{纬}=\text{constant}$

Here $P$is the pressure of air, $V$is the volume of the air, and $纬$is the adiabatic constant.

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

$P_i{V_i}^{纬}=P_f{V_f}^{纬}$

Equation (2) can be simplified as

${V_f}^{纬}=\frac{P_f{V_f}^{纬}}{P_f}$

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

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

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

Hence, the required final volume is $\text{2.49}脳{\text{10}}^{\text{-4}}{\text{鈥尘}}^{\text{3}}$.

1 liter of air is compressed adiabatically to a pressure of 7 atm. The initial volume of air is$1脳{10}^{鈭�3}{\text{鈥尘}}^{\text{3}}$ and the final volume is $2.49脳{10}^{鈭�4}{\text{鈥尘}}^{\text{3}}$.

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, $V_i\text{鈥塧苍诲鈥�}{V}_f$are initial volume, and final volume.

Expression for the adiabatic process is

role="math" localid="1651473925139" $P{V}^{纬}=\text{const}$ 鈥︹�� (2)

Here $P$is the pressure of air,$V$ is the volume of the air, and$纬$ is the adiabatic constant.

Equation (2) can be written as

$T{V}^{纬鈭�1}=\text{const}$

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

$P=c{V}^{鈭�纬}$ 鈥︹�� (3)

Substitute $c{V}^{鈭�纬}$for $P$in equation (1)

$W=鈭�c\underset{1脳{10}^{鈭�3}}{\overset{2.49脳{10}^{鈭�4}}{鈭�}}{V}^{鈭�纬}dV$ 鈥︹�� (4)

Solving equation (4) for$W$

$W=鈭�c{\left[\frac{V^{鈭�纬+1}}{鈭�纬+1}\right]}_{1脳{10}^{鈭�3}}^{2.49脳{10}^{鈭�4}}$ 鈥︹�� (5)

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

$W=\mathrm{29.48.}c$ 鈥︹�� (6)

At $V=1\text{鈥塴颈迟别谤}$=$1脳{10}^{鈭�3}{\text{鈥尘}}^{\text{3}}$and role="math" localid="1651475635356" $101325\text{鈥塒补}$for $P$in equation (2)

$\begin{array}{l}P{V}^{纬}=c\\ 1.013脳{10}^5脳{\left(1脳{10}^{鈭�3}\right)}^{\frac{7}{5}}=c\\ c=6.393\text{鈥塒补.}\end{array}$

Substitute for in equation (6)

$\begin{array}{l}W=29.48脳6.393\\ \boxed{W=188.44\text{鈥塉}}\end{array}$

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

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

The initial volume is$V_i=1脳{10}^{鈭�3}{\text{鈥尘}}^{\text{3}}$ and the final volume is $V_f=2.49脳{10}^{鈭�4}{\text{鈥尘}}^{\text{3}}$.

Expression for adiabatic compression is

$V{T}^{\left(\frac{f}{2}\right)}=\text{const}$ 鈥︹�� (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

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

鈥︹�� (3)

Substitute 5 for $f$, $1脳{10}^{鈭�3}{\text{m}}^{\text{3}}$for $V_i$, $2.49脳{10}^{鈭�4}{\text{鈥尘}}^{\text{3}}$for $V_f$on equation (3)

$\begin{array}{l}{T}_f={\left(\frac{1脳{10}^{鈭�3}}{2.49脳{10}^{鈭�4}}\right)}^{\frac{2}{5}}脳300\\ \boxed{T_f=523.17\text{鈥块}}\end{array}$

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