Question

A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Take the initial pressure and temperature of the air before it is put in the tire to be Pi = 1.00 bar and Ti = 298 K. The final volume of the air in the tire is Vf= 1.50 L and the final pressure is Pf = 5.00 bar. Calculate the final temperature of the air in the tire.

Answer 1

Complete Question

A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Take the initial pressure and temperature of the air before it is put in the tire to be Pi = 1.00 bar and Ti = 298 K. The final volume of the air in the tire is Vf= 1.50 L and the final pressure is Pf = 5.00 bar. Calculate the final temperature of the air in the tire.

Assume that Cv,m = 5R/2

Answer:

The value is $T_f = 471.978 \ K$

Explanation:

From the question we are told that

The initial pressure is $P_i = 1.00 \ bar$

The initial temperature is $T_i = 298 K$

The final volume is $V_f = 1.50 \ L$

The final pressure is $P_f = 5.0 \ bar$

Generally the equation for reversible adiabatic compression is

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

Generally from the ideal gas equation we have that

$PV = n RT$

=> $V = \frac{n RT}{P}$

So

$P_i *[ \frac{n RT_i}{P_i}]^{\gamma} = P_f [ \frac{n RT_f}{P_f}]^{\gamma}$

=> $P_i ^{1 - \gamma} * T_i ^{\gamma} = P_f ^{1-\gamma} * T_f^{\gamma}$

=> $T_f ^{\gamma} = \frac{ P^{1 - \gamma * T_i^{\gamma}}}{ P_f ^{1 - \gamma}}$

=> $T_f =[ \frac{ P_i ^{1 - \gamma} * T_i^{\gamma}}{P_f^{1 - \gamma}}]^{\frac{1}{\gamma} }$

=> $T_f = T_i * [\frac{P_i}{P_f} ]^{\frac{1- \gamma}{\gamma}$

Here $\gamma$ is a constant mathematically represented as

$\gamma = \frac{C_{P,m}}{C_{V, m}}$

Here

$C_{P,m}}$ is the molar heat capacity at constant pressure

and $C_{V,m}}$ is the molar heat capacity at constant volume given as

$C_{V,m}} = \frac{5R}{2}$

Generally $C_{P,m}}$ for an ideal gas is mathematically represented as

$C_{P,m}} = R +C_{V,m}}$

So

$\gamma = \frac{\frac{5R}{2} + R}{ \frac{5R}{2} }$

=>   $\gamma = \frac{7}{5}$

So

=> $T_f = T_i * [\frac{P_i}{P_f} ]^{\frac{1- \frac{7}{5}}{ \frac{7}{5}}$

=> $T_f = 298 * [\frac{1}{5} ]^{\frac{1- \frac{7}{5}}{ \frac{7}{5}}$

=> $T_f = 471.978 \ K$