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Q.4.33

Expert-verified
Found in: Page 143

### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

# Table 4.5 gives experimental values of the molar enthalpy of nitrogen at 1 bar and 100 bars. Use this data to answer the following questions about a nitrogen throttling process operating between these two pressures.(a) If the initial temperature is $300\mathrm{K}$, what is the final temperature? (Hint: You'll have to do an interpolation between the tabulated values.)(b) If the initial temperature is $200\mathrm{K}$, what is the final temperature?(c) If the initial temperature is $100\mathrm{K}$, what is the final temperature? What fraction of the nitrogen ends up as a liquid in this case?(d) What is the highest initial temperature at which some liquefaction takes place?(e) What would happen if the initial temperature were $600\mathrm{K}$? Explain.

a) The final temperature is $281.4k$

b) The final temperature is $153.9K$

c) The final temperature after the throttling process is $77\mathrm{K}$. Thus, the fraction of nitrogen that becomes liquefied is $\underset{¯}{0.7376}$.

d)role="math" localid="1648594282162" $\text{The highest initial temperature at which some liquefaction takes place is}\phantom{\rule{0ex}{0ex}}164.3\mathrm{K}$

e) role="math" localid="1648594270697" $\text{The final temperature increases when the initial temperature is}600\mathrm{K}\text{.}$

See the step by step solution

## Part (a) - Step 1: To find

The final temperature

## Part (a) - Step 2: Explanation

Given: A nitrogen throttling process operates between pressures 1 bar and 100 bars. The initial temperature is $300\mathrm{K}$.

Formula: The value of the initial enthalpy at temperature $300\mathrm{K}$ and pressure 100 bars is $8174\mathbf{J}$, the enthalpy at temperature $300\mathrm{K}$ and pressure 1 bar is $8717\mathrm{J}$, and the enthalpy at temperature $200\mathrm{K}$ and pressure 1 bar is $5800\mathrm{J}$ from table 4.5

The liquid nitrogen fraction x has the following value:

$x=\frac{\left(8174\mathrm{J}\right)-\left(8717\mathrm{J}\right)}{\left(5800\mathrm{J}\right)-\left(8717\mathrm{J}\right)}\phantom{\rule{0ex}{0ex}}=0.186$

$\text{The expression of the final temperature}{T}_{\mathrm{F}}$

${T}_{\mathrm{F}}=\left(200\mathrm{K}\right)x+\left(300\mathrm{K}\right)\left(1-x\right)\dots \left(1\right)$

Calculation: Substitute $x=0.186inequation\left(1\right)$

${T}_{\mathrm{F}}=\left(200\mathrm{K}\right)\left(0.186\right)+\left(300\mathrm{K}\right)\left(1-0.186\right)\phantom{\rule{0ex}{0ex}}=281.4k$

Hence the final temperature is $281.4k$

## Part (b) - Step 3: To find

The final temperature

## Part (b) - Step 4: Explanation

Given:

A nitrogen throttling process operates between pressures 1 bar and 100 bars. The initial temperature is $200\mathrm{K}$.

Formula : The value of the initial enthalpy at temperature $200\mathrm{K}$ and pressure 100 bars is $4442\mathrm{J}$, the enthalpy at temperature $200\mathrm{K}$and pressure $1\mathrm{bar}$ is $5800\mathrm{J}$, and the enthalpy at temperature $100\mathrm{K}$ and pressure 1 bar is $2856\mathrm{J}$from table 4.5.

The fraction of nitrogen x that becomes liquefied has the following value:

$x=\frac{\left(4442\mathrm{J}\right)-\left(5800\mathrm{J}\right)}{\left(2856\mathrm{J}\right)-\left(5800\mathrm{J}\right)}\phantom{\rule{0ex}{0ex}}=0.461$

The expression of the final temperature ${T}_{\mathrm{F}}$

${T}_{\mathrm{F}}=\left(100\mathrm{K}\right)x+\left(200\mathrm{K}\right)\left(1-x\right)\dots .\left(2\right)$

Substitute X= 0.461 in equation (2)

${T}_{\mathrm{F}}=\left(100\mathrm{K}\right)\left(0.461\right)+\left(200\mathrm{K}\right)\left(1-0.461\right)\phantom{\rule{0ex}{0ex}}=153.9\mathrm{K}$

Hence the final temperature is 153.9K

## Part (c) - Step 5: To find

The final temperature. and the fraction of nitrogen that becomes liquefied.

## Part(c) - Step 6: Explanation

Given:

A nitrogen throttling process operates between pressures 1 bar and 100 bars. The initial temperature is $100\mathrm{K}$.

Formula: In order to conserve the initial enthalpy, the temperature has to decrease. Since the decreased temperature lies between the enthalpies for gaseous and liquid nitrogen at pressure 1 bar and temperature $77\mathrm{K}$, therefore, the temperature will be$77\mathrm{K}$after the throttling process.

$\text{The expression of the initial enthalpy}H\text{at temperature}100\mathrm{K}\phantom{\rule{0ex}{0ex}}\text{and pressure}100\text{bars}$

$H={H}_{\mathrm{liq}}x+{H}_{\mathrm{gas}}\left(1-x\right)$

Here, x is fraction of liquefaction, ${H}_{\text{liq}}$is the enthalpy of liquefied nitrogen and ${H}_{\text{gas}}$is the enthalpy of gaseous nitrogen at pressure$1\mathrm{bar}$ and temperature $77\mathrm{K}$.

Rearrange the above expression for $x$

$x=\frac{H-{H}_{gE}}{{H}_{\mathrm{fiq}}-{H}_{gas}}\dots \text{(3)}$

Calculation:Substitute H= $-1946\mathrm{J}$ ${H}_{\text{liq}}$ =$-3407J$androle="math" localid="1648599827773" ${H}_{\text{gas}}=2161\mathrm{J}$ in equation (3)

$x=\frac{\left(-1946\mathrm{J}\right)-\left(2161\mathrm{J}\right)}{\left(-3407\mathrm{J}\right)-\left(2161\mathrm{J}\right)}\phantom{\rule{0ex}{0ex}}=0.7376$

Thus, the final temperature after the throttling process is $77\mathrm{K}$.

Thus, the fraction of nitrogen that becomes liquefied is $\underset{¯}{0.7376}$.

## Part (d) - Step 7: To find

The highest initial temperature at which some liquefaction takes place.

## Part(d) - Step 8: Explanation

Given: $\text{A nitrogen throttling process operates between pressures}1\text{bar and}\phantom{\rule{0ex}{0ex}}100\text{bars.}$

Formula: The temperature at which liquefaction occurs is when the enthalpy of the liquid equals the enthalpy of the gas. It's somewhere between$100\mathrm{K}$and $200\mathrm{K}$in temperature.

The value of the enthalpy of gas is $2161\mathrm{J}$, the enthalpy at temperature $100\mathrm{K}$ is $-1946\mathrm{J}$, and the enthalpy at temperature$200\mathrm{K}$ is $4442\mathrm{J}$ from table 4.5.

$\text{The value of the fraction of nitrogen}x\text{that becomes liquefied is:}$

$x=\frac{\left(2161\mathrm{J}\right)-\left(4442\mathrm{J}\right)}{\left(-1946\mathrm{J}\right)-\left(4442\mathrm{J}\right)}\phantom{\rule{0ex}{0ex}}=0.357$

The expression of the highest temperature ${T}_{\mathrm{H}}$

${T}_{\mathrm{H}}=\left(100\mathrm{K}\right)x+\left(200\mathrm{K}\right)\left(1-x\right)\dots \left(4\right)$

Calculation: Substitute x= 0.357 in equation (4)

${T}_{\mathrm{F}}=\left(100\mathrm{K}\right)\left(0.357\right)+\left(200\mathrm{K}\right)\left(1-0.357\right)\phantom{\rule{0ex}{0ex}}=164.3\mathrm{K}$

$\text{Hence the highest initial temperature at which some liquefaction}\text{t}\phantom{\rule{0ex}{0ex}}\text{akes place is}164.3\mathrm{K}$

## Part (e) - Step 9: To find

When the initial temperature is $600\mathrm{K}$ the following is the result.

## Part (e) - Step 10: Explanation

The nitrogen throttling technique works at pressures ranging from 1 bar to 100 bar.

Because the enthalpy at a pressure of 100 bars is greater than the enthalpy at a pressure of 1 bar when the initial temperature is$600\mathrm{K}$ the temperature after the throttling operation will rise rather than decrease, in contrast to earlier parts.

$\text{Hence the final temperature increases when the initial temperature is}\phantom{\rule{0ex}{0ex}}600\mathrm{K}.$