Question

A tank contains helium at \(37^{\circ} \mathrm{C}\) and 140 kPa gage. The helium is heated in a process by heat transfer from the surroundings such that the helium reaches a final equilibrium state at \(200^{\circ} \mathrm{C}\). Determine the final gage pressure of the helium. Assume atmospheric pressure is \(100 \mathrm{kPa}\)

Short answer

Short Answer: The final gage pressure of the helium in the tank when heated to 200°C is 265.40 kPa.

Step-by-step solution

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin to use them in our calculations. The conversion formula is K = C + 273.15. Initial temperature (\(T_1\)): \(37^{\circ} \mathrm{C} = 37 + 273.15 = 310.15 \mathrm{K}\) Final temperature (\(T_2\)): \(200^{\circ} \mathrm{C} = 200 + 273.15 = 473.15 \mathrm{K}\)

Determine the initial and final absolute pressures

Now, we need to determine the initial and final absolute pressures. The gage pressure is above atmospheric pressure, so we add the atmospheric pressure to get the absolute pressure (\(P_{abs}\)). Initial gage pressure (\(P_{gage1}\)): 140 kPa Initial absolute pressure (\(P_1\)): \(P_{gage1} + P_{atm} = 140\,\text{kPa} + 100\,\text{kPa} = 240\,\text{kPa}\) The final absolute pressure (\(P_2\)) is what we need to find.

Apply the ideal gas law

The ideal gas law relating pressure, volume, temperature, and the gas constant is given as: \(pV = nRT\) Since the process is isobaric (constant volume) and we are concerned only about the pressures and temperatures, we can write the ideal gas law for both initial and final states as follows: \(P_1V = nR T_1\) and \(P_2V = nRT_2\) Where \(P_1\) and \(P_2\) are the initial and final absolute pressures, \(V\) is the volume, \(n\) is the number of moles of helium, \(R\) is the gas constant, and \(T_1\) and \(T_2\) are the initial and final temperatures in Kelvin.

Solve for the final gage pressure

Dividing the second equation by the first equation, we get: \(\dfrac{P_2V}{P_1V} = \dfrac{nRT_2}{nRT_1}\) \(\dfrac{P_2}{P_1} = \dfrac{T_2}{T_1}\) Now plug in the values of \(P_1\), \(T_1\), and \(T_2\) to find the final absolute pressure (\(P_2\)). \(\dfrac{P_2}{240\,\text{kPa}} = \dfrac{473.15\,\text{K}}{310.15\,\text{K}}\) Solving for \(P_2\): \(P_2 = 240 \times \dfrac{473.15}{310.15} = 365.40 \,\text{kPa}\) Finally, to find the final gage pressure (\(P_{gage2}\)), we subtract the atmospheric pressure from the final absolute pressure (\(P_2\)). \(P_{gage2} = P_2 - P_{atm} = 365.40\,\text{kPa} - 100\,\text{kPa} = 265.40\,\text{kPa}\) The final gage pressure of the helium after heating is \(265.40\,\text{kPa}\).