Question

Nitrogen gas to 20 psia and \(100^{\circ} \mathrm{F}\) initially occupies a volume of \(1 \mathrm{ft}^{3}\) in a rigid container equipped with a stirring paddle wheel. After 5000 lbf.ft of paddle wheel work is done on nitrogen, what is its final temperature?

Short answer

Answer: The final temperature of Nitrogen gas is 459.55°F.

Step-by-step solution

Write down the given information

Initial pressure of Nitrogen gas, \(P_1 = 20\) psia Initial temperature, \(T_1 = 100^\circ \mathrm{F}\) Initial volume, \(V_1 = 1 \mathrm{ft}^3\) Work done on Nitrogen gas, \(W = 5000\) lbf.ft

Convert the temperature to Rankine and pressure to lbf/ft²

First, we need to convert the initial temperature from Fahrenheit to Rankine and pressure from psia to lbf/ft². Temperature in Rankine, \(T_1 = 100 + 459.67 = 559.67\,^\circ \mathrm{R}\) Pressure in lbf/ft², \(P_1 = 20 \times 144 = 2880 \mathrm{lbf/ft^2}\) \ Since volume does not change, \(V_2 = V_1 = 1\,\mathrm{ft^3}\)

Use the Ideal Gas Law to find the initial mass of Nitrogen gas

We can use the Ideal Gas Law to find the initial mass (m) of Nitrogen gas. The properties of Nitrogen gas are given as follows: \ $$ PV = mRT \\ m = \frac{PV}{RT} $$ Gas constant, \(R_{nitrogen} = 55.15 \frac{\mathrm{lbf.ft}}{\mathrm{lbm.R}}\) $$ m = \frac{2880 \times 1}{55.15 \times 559.67} = 0.0937\,\mathrm{lbm} $$

Apply the First Law of Thermodynamics to the system

Since the volume does not change, the work done on the gas causes an increase in internal energy. The First Law of Thermodynamics states that the change in internal energy (\(\Delta U\)) is equal to the work done on a system. Hence, $$ \Delta U = W $$

Relate the change in internal energy to the change in temperature

The change in internal energy (\(\Delta U\)) can also be related to specific heat capacity at constant volume (\(c_v\)) and the change in temperature (\(\Delta T = T_2 - T_1\)) as follows: $$ \Delta U = mc_v \Delta T \\ $$ The specific heat capacity at constant volume for Nitrogen gas is \(c_v = 0.171\,\frac{\mathrm{Btu}}{\mathrm{lbm.R}}\). Converting Btu to lbf.ft, \(c_v = 0.171 \times 778.16 = 133.036\,\frac{\mathrm{lbf.ft}}{\mathrm{lbm.R}}\) By combining the two equations, we get $$ mc_v \Delta T = W \\ \Delta T = \frac{W}{mc_v} $$

Calculate the final temperature and convert back to Fahrenheit

Finally, we can calculate the change in temperature and then the final temperature: $$ \Delta T = \frac{5000}{0.0937 \times 133.036} = 359.55\,^\circ \mathrm{R} $$ Now, we can find the final temperature in Rankine and convert it back to Fahrenheit: $$ T_2 = T_1 + \Delta T = 559.67 + 359.55 = 919.22\,^\circ \mathrm{R} \\ T_{2(Fahrenheit)} = 919.22 - 459.67 = 459.55^\circ \mathrm{F} $$ Thus, the final temperature of Nitrogen gas is \(459.55^\circ \mathrm{F}\).