Adiabatic Processes for an Ideal Gas Problems and Solutions

Q#19.29

A monatomic ideal gas that is initially at a pressure of 1.50 $\times 10^5$ Pa and has a volume of 0.0800 $m^3$ is compressed adiabatically to a volume of 0.0400 $m^3$

(a) What is the final pressure? 

(b) How much work is done by the gas? 

(c) What is the ratio of the final temperature of the gas to its initial temperature? Is the gas heated or cooled by this compression? 

Answer:

For an adiabatic process of an ideal gas,

$p_1V_1^{\gamma}$ = $p_2V_2^{\gamma}$ (*)

W = $\frac{1}{\gamma-1}(p_1V_1-p_2V_2)$ (**) and

$T_1V_1^{\gamma-1}$ = $T_2V_2^{\gamma-1}$ (***)

For a monatomic ideal gas $\gamma = \frac{5}{3}$

(a) $p_1=p_2\left(\frac{V_1}{V_2}\right)^{\gamma}$

$p_1=(1.50 \times 10^5 \ Pa)\left(\frac{0.0800 \ m^3}{0.0400 \ m^3}\right)^{5/3}$

$p_1=4.76 \times 10^5$ Pa

(b) This result may be substituted into Eq. (*), or, substituting the above form for $p_2$,

W = $\frac{1}{\gamma-1}\left[p_1V_1-p_1\left(\frac{V_1}{V_2}\right)^{\gamma}V_2\right]$

W = $\frac{1}{\gamma-1}\left[p_1V_1-p_1V_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}\right]$

W = $\frac{p_1V_1}{\gamma-1}\left[1-\left(\frac{0.0800 \ m^3}{0.0400 \ m^3}\right)^{5/3-1}\right]$

W = $\frac{3}{2}(1.50 \times 10^5 \ Pa)(0.0800 \ m^3)\left[1-\left(\frac{V_1}{V_2}\right)^{\gamma-1}\right]$

W = $-1.06 \times 10^4$ J

(c) From (***) we write

$\frac{T_1}{T_2}=\left(\frac{V_1}{V_2}\right)^{\gamma -1}$

$\frac{T_2}{T_1}=\left(\frac{0.0800 \ m^3}{0.0400 \ m^3}\right)^{5/3-1}=1.59$

and since the final temperature is higher than the initial temperature, the gas is heated.

Q#19.30

In an adiabatic process for an ideal gas, the pressure decreases. In this process does the internal energy of the gas increase or decrease? Explain your reasoning. 

Answer:

For an ideal gas ΔU = n$C_v\Delta T$ The sign of ΔU is the same as the sign of ΔT. Combine Eq. (***) and the ideal gas law  to obtain an equation relating T and p, and use it to determine the sign of ΔT.

$T_1V_1^{\gamma-1}$ = $T_2V_2^{\gamma-1}$ 

and pV = nRT → $V = \frac{nRT}{p}$

$T_1\left(\frac{nRT_1}{p_1}\right)^{\gamma-1} = T_2\left(\frac{nRT_2}{p_2}\right)^{\gamma-1}$

$T_1\left(\frac{T_1}{p_1}\right)^{\gamma-1} = T_2\left(\frac{T_2}{p_2}\right)^{\gamma-1}$

$\frac{T_1^{\gamma}}{p_1^{\gamma - 1}} = \frac{T_2^{\gamma}}{p_2^{\gamma -1}}$

$p_2$ < $p_1$ and γ − 1 is positive so $T_2$ < $T_1$. ΔT is negative so ΔU is negative; the energy of the gas decreases. 

Eq. (*) shows that the volume increases for this process, so it is an adiabatic expansion. In an adiabatic expansion the temperature decreases. 

Q#19.31

Two moles of carbon monoxide (CO) start at a pressure of 1.2 atm and a volume of 30 liters. The gas is then compressed adiabatically to $\frac{1}{3}$ this volume. Assume that the gas may be treated as ideal. What is the change in the internal energy of the gas? Does the internal energy increase or decrease? Does the temperature of the gas increase or decrease during this process? Explain

Answer:

For an adiabatic process of an ideal gas,  W = $\frac{1}{\gamma-1}(p_1V_1-p_2V_2)$ and

$p_1V_1^{\gamma}$ = $p_2V_2^{\gamma}$

Given: $p_1=1.22 \times 10^5$ Pa

 γ = 1.40 for an ideal diatomic gas. 

1 L = $1 \times 10^{-3} \ m^3$ and 1 atm = 1.013 $\times 10^5$ Pa

Q = ΔU + W = 0 for an adiabatic process, so $\Delta U = -W$

$\Delta U = -\frac{1}{\gamma-1}(p_1V_1-p_2V_2)$

with $p_2=p_1\left(\frac{V_1}{V_2}\right)^{\gamma}$

$p_2=(1.22 \times 10^5 \ Pa)\left(\frac{V_1}{\frac{1}{3}V_1}\right)^{1.4}$

$p_2=5.68 \times 10^5$ Pa

So, $\Delta U = \frac{1}{1.4-1}[(5.68 \times 10^5 \ Pa)(10 \times 10^{-3} \ m^3)-(1.22 \times 10^{-3} \ m^3)(30 \times 10^{-3} \ m^3)]$

$\Delta U = 5.05 \times 10^{3}$ J

energy increases because work is done on the gas (ΔU > 0) and Q = 0. The temperature increases because the internal energy has increased. 

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