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Chapter 15: Problem 9

Suppose 1.00 mol of oxygen is heated at constant pressure of 1.00 atm from \(10.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C} .\) (a) How much heat is absorbed by the gas? (b) Using the ideal gas law, calculate the change of volume of the gas in this process. (c) What is the work done by the gas during this expansion? (d) From the first law, calculate the change of internal energy of the gas in this process.

Short Answer

Expert verified
Question: A 1.00 mol oxygen gas (O2) is heated from 10.0°C to 25.0°C under constant pressure of 1.00 atm. The heat capacity of O2 under constant pressure (Cp) is 29.4 J/mol.K. Calculate (a) the heat absorbed by the gas, (b) the change in volume, (c) the work done during expansion, and (d) the change in internal energy. Answer: (a) The heat absorbed by the gas is 441 J. (b) The change in volume is 0.00206 m^3. (c) The work done during expansion is -208.73 J. (d) The change in internal energy is 232.27 J.

Step by step solution

01

Part (a): Calculate the heat absorbed by the gas

To find the heat absorbed (q) by the gas, use the formula: q = n * Cp * ΔT where n is the number of moles, Cp is the heat capacity at constant pressure, and ΔT is the change in temperature (T2 - T1). ΔT = T2 - T1 = 25.0 - 10.0 = 15.0 K q = (1.00 mol) * (29.4 J/mol.K) * (15.0 K) q = 441 J So, the gas absorbs 441 J of heat.
02

Part (b): Calculate the change in volume

We can use the ideal gas law to find the change in volume (ΔV): PV = nRT First, convert pressure from atm to Pa by multiplying it with the conversion factor (1 atm = 101325 Pa). Then, calculate the initial and final volumes using the ideal gas law: P = 1.00 atm * 101325 Pa/atm = 101325 Pa R = 8.314 J/mol.K Initial volume (V1): V1 = nRT1 / P V1 = (1.00 mol)(8.314 J/mol.K)(283.15 K) / 101325 Pa V1 = 0.02241 m^3 Final volume (V2): V2 = nRT2 / P V2 = (1.00 mol)(8.314 J/mol.K)(298.15 K) / 101325 Pa V2 = 0.02447 m^3 Change in volume (ΔV): ΔV = V2 - V1 ΔV = 0.02447 - 0.02241 ΔV = 0.00206 m^3 The change in volume is 0.00206 m^3.
03

Part (c): Calculate the work done

To find the work done (w) by the gas during expansion, we use the formula: w = -P(ΔV) w = -101325 Pa * 0.00206 m^3 w = -208.73 J The work done by the gas during expansion is -208.73 J.
04

Part (d): Calculate the change in internal energy

To find the change in internal energy (ΔU), we use the first law of thermodynamics: ΔU = q + w ΔU = 441 J + (-208.73 J) ΔU = 232.27 J The change in internal energy of the gas is 232.27 J.

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