Question

One mole of oxygen is heated at constant pressure starting at $0^{\circ} \mathrm{C}$. How much heat energy in cal must be added to the gas to double its volume ? Take \(\mathrm{R}=2\\{(\mathrm{cal}) /(\) mole \()\\}\) (A) 1938 (B) 1920 (C) 1911 (D) 1957

Short answer

The amount of heat energy required to double the volume of one mole of oxygen gas at constant pressure, starting at 0°C, is 1911 cal.

Step-by-step solution

Identify the relevant formula for constant pressure heating.

For constant pressure heating, we can use the formula: ΔQ = nCpΔT where ΔQ is the heat energy added, n is the number of moles, Cp is the specific heat at constant pressure, and ΔT is the change in temperature.

Relate the initial and final volume using the ideal gas law.

The initial and final conditions of the gas can be related using the ideal gas law: PV = nRT Since the pressure is constant, we can write the following equation for the initial and final states: \(V_{1} / T_{1} = V_{2} / T_{2}\) We are given that the volume doubles (V2 = 2V1) and that the initial temperature is 0°C, which is equivalent to 273 K. We can now find ΔT: \(V_{1} / 273 K = 2V_{1} / T_{2}\)

Solve for the final temperature T2.

Solving the equation from step 2 for T2, we get: \(T_{2} = 2 * 273 K = 546 K\) Now that we have the initial and final temperatures, we can find ΔT: ΔT = T2 - T1 = 546 K - 273 K = 273 K

Find the specific heat at constant pressure, Cp.

The specific heat at constant pressure for a diatomic gas like oxygen can be calculated using the formula: Cp = (7/2)R Given that R = 2 cal/(mole), Cp can be calculated as: Cp = (7/2) * 2 cal/(mole) = 7 cal/(mole)

Calculate the heat energy added, ΔQ.

Now, we can use the constant pressure heating formula from step 1 with n = 1 mole, Cp = 7 cal/(mole), and ΔT = 273 K: ΔQ = nCpΔT = (1 mole) * (7 cal/mole) * (273 K) = 1911 cal Therefore, 1911 cal of heat energy must be added to the system to double the volume, which corresponds to choice (C).