Question

An ideal gas is heated at a constant pressure of $2.0 \times 10^{5} \mathrm{Pa}\( from a temperature of \)-73^{\circ} \mathrm{C}$ to a temperature of \(+27^{\circ} \mathrm{C}\). The initial volume of the gas is $0.10 \mathrm{m}^{3} .\( The heat energy supplied to the gas in this process is \)25 \mathrm{kJ} .$ What is the increase in internal energy of the gas?

Short answer

Answer: The increase in internal energy of the gas is 15 kJ.

Step-by-step solution

Convert temperatures to Kelvin

To do any calculations with temperatures, it's essential to convert them to Kelvin first. To convert a Celsius temperature to Kelvin, we simply add 273.15. Initial temperature in Kelvin: \(T_1 = -73^{\circ}\mathrm{C} + 273.15 = 200.15\ \mathrm{K}\) Final temperature in Kelvin: \(T_2 = 27^{\circ}\mathrm{C} + 273.15 = 300.15\ \mathrm{K}\)

Calculate the final volume of the gas

Since we are given the constant pressure, we can use the ideal gas law to find the final volume of the gas. The ideal gas law can be written as \(PV = nRT\) Since the pressure, number of moles n, and the gas constant R are constant, we can write \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) We are given the initial volume (\(V_1 = 0.10\ \mathrm{m}^3\)) and have found the initial and final temperatures in Kelvin. Now, we can solve for the final volume \(V_2\): \(V_2 = \frac{T_2}{T_1} \times V_1 = \frac{300.15\ \mathrm{K}}{200.15\ \mathrm{K}} \times 0.10\ \mathrm{m}^3 \approx 0.150\ \mathrm{m}^3\)

Calculate the work done by the gas

The work done by the gas can be calculated using the constant pressure and the change in volume: \(W = P \Delta V = P(V_2 - V_1) = (2.0 \times 10^5\ \mathrm{Pa}) (0.150\ \mathrm{m}^3 - 0.10\ \mathrm{m}^3) = 10\ \mathrm{kJ}\)

Use the First Law of Thermodynamics to find the increase in internal energy

The First Law of Thermodynamics states that \(\Delta U = Q - W\) Where \(\Delta U\) is the change in internal energy, \(Q\) is the heat added, and \(W\) is the work done by the system. We are given the amount of heat energy supplied to the gas \(Q = 25\ \mathrm{kJ}\), and we have calculated the work done by the gas \(W = 10\ \mathrm{kJ}\). Now we can find the increase in internal energy: \(\Delta U = 25\ \mathrm{kJ} - 10\ \mathrm{kJ} = 15\ \mathrm{kJ}\) So, the increase in internal energy of the gas is 15 kJ.