One mole of an ideal gas expands isothermally and reversibly at 0°C. The pressure on 1 mol of an ideal monatomic gas changes from 100.0 atm to 1.00 atm.
-Calculate q.

Question

Quizplus · Accepted Answer

Since the gas expands isothermally, the temperature remains constant at 0°C or 273 K.

From the ideal gas law, PV = nRT, we can write:

P1V1 = nRT, where P1 = 100.0 atm, V1 = V, n = 1 mol, and R is the gas constant.

For an isothermal process, we have:

P1V1 = P2V2, where P2 = 1.00 atm and V2 = 22.4 L/mol (since the gas is monatomic and we are at standard temperature and pressure).

Therefore, we can calculate V1 as:

V1 = (P2/P1) * V2 = (1.00 atm/100.0 atm) * 22.4 L/mol = 0.224 L/mol

The work done (W) by the gas is given by:

W = -PΔV = -P1(V2 - V1) = -100.0 atm * (0.224 L/mol - 22.4 L/mol) = 2.24 kJ/mol

Since the process is reversible and isothermal, the change in internal energy (ΔU) is zero. Therefore, the heat transferred (q) is equal to the work done:

q = -W = -2.24 kJ/mol = -10.5 kJ (rounded to one decimal place) for 1 mol of gas.

Therefore, the correct answer is D) -10.5 kJ.

One Mole of an Ideal Gas Expands Isothermally and Reversibly