Question

Problem 1.36. In the course of pumping up a bicycle tire, a liter of air at atmospheric pressure is compressed adiabatically to a pressure of 7 atm. (Air is mostly diatomic nitrogen and oxygen.)

(a) What is the final volume of this air after compression?

(b) How much work is done in compressing the air?

(c) If the temperature of the air is initially$300\text{K}$ , what is the temperature after compression?

Short answer

Part (a) The final volume of the air after compression is $\boxed{V_f=\text{2.49}\times {\text{10}}^{\text{-4}}{\text{\hspace{0.17em}m}}^{\text{3}}}$.

Part (b) Work done on the air is $\boxed{W=188.44\text{\hspace{0.17em}J}}$.

Part (c) The final temperature after compression is $\boxed{T_f=523.17\text{\hspace{0.17em}K}}$.

Step-by-step solution

Part a. Step 1. Given information.

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

Air is mostly diatomic.

Part a. Step 2. Explanation.

Expression for the adiabatic process is

$P{V}^{\gamma }=\text{constant}$

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

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

$P_i{V_i}^{\gamma }=P_f{V_f}^{\gamma }$

Equation (2) can be simplified as

${V_f}^{\gamma }=\frac{P_f{V_f}^{\gamma }}{P_f}$

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

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

$\begin{array}{l}{V}_f={\left(\frac{1}{7}\right)}^{\frac{5}{7}}\times 1\\ V_f=0.249\text{\hspace{0.17em}L=2.49}\times {\text{10}}^{\text{-4}}{\text{\hspace{0.17em}m}}^{\text{3}}\\ \boxed{V_f=\text{2.49}\times {\text{10}}^{\text{-4}}{\text{\hspace{0.17em}m}}^{\text{3}}}\end{array}$

Hence, the required final volume is $\text{2.49}\times {\text{10}}^{\text{-4}}{\text{\hspace{0.17em}m}}^{\text{3}}$.

Part b. Step 1. Given information.

1 liter of air is compressed adiabatically to a pressure of 7 atm. The initial volume of air is$1\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}}$ and the final volume is $2.49\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{3}}$.

Part b. Step 2. Explanation.

Expression for work done on the system is

$W=-\underset{v_i}{\overset{v_f}{\int }}PdV$ …… (1)

Here $P$is the pressure, $V_i\text{\hspace{0.17em}and\hspace{0.17em}}{V}_f$are initial volume, and final volume.

Expression for the adiabatic process is

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

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

Equation (2) can be written as

$T{V}^{\gamma -1}=\text{const}$

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

$P=c{V}^{-\gamma }$ …… (3)

Substitute $c{V}^{-\gamma }$for $P$in equation (1)

$W=-c\underset{1\times {10}^{-3}}{\overset{2.49\times {10}^{-4}}{\int }}{V}^{-\gamma }dV$ …… (4)

Solving equation (4) for$W$

$W=-c{\left[\frac{V^{-\gamma +1}}{-\gamma +1}\right]}_{1\times {10}^{-3}}^{2.49\times {10}^{-4}}$ …… (5)

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

$W=\mathrm{29.48.}c$ …… (6)

At $V=1\text{\hspace{0.17em}liter}$=$1\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}}$and role="math" localid="1651475635356" $101325\text{\hspace{0.17em}Pa}$for $P$in equation (2)

$\begin{array}{l}P{V}^{\gamma }=c\\ 1.013\times {10}^5\times {\left(1\times {10}^{-3}\right)}^{\frac{7}{5}}=c\\ c=6.393\text{\hspace{0.17em}Pa.}\end{array}$

Substitute for in equation (6)

$\begin{array}{l}W=29.48\times 6.393\\ \boxed{W=188.44\text{\hspace{0.17em}J}}\end{array}$

Hence, work done on the air is $\boxed{W=188.44\text{\hspace{0.17em}J}}$.

Part c. Step 1. Given information.

The initial temperature is $T_i=300\text{\hspace{0.17em}K}$.

The initial volume is$V_i=1\times {10}^{-3}{\text{\hspace{0.17em}m}}^{\text{3}}$ and the final volume is $V_f=2.49\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{3}}$.

Part c. Step 2. Explanation.

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 $f$, $1\times {10}^{-3}{\text{m}}^{\text{3}}$for $V_i$, $2.49\times {10}^{-4}{\text{\hspace{0.17em}m}}^{\text{3}}$for $V_f$on equation (3)

$\begin{array}{l}{T}_f={\left(\frac{1\times {10}^{-3}}{2.49\times {10}^{-4}}\right)}^{\frac{2}{5}}\times 300\\ \boxed{T_f=523.17\text{\hspace{0.17em}K}}\end{array}$

Hence, the final temperature after compression is $\boxed{T_f=523.17\text{\hspace{0.17em}K}}$.