One mole of an ideal gas initially at a temperature of Ti = 1.0°C undergoes an expansion at a constant pressure of 1.00 atm to nine times its original volume.(a) Calculate the new temperature Tf of the gas. K(b) Calculate the work done on the gas during the expansion. kJ

(a) To calculate the new temperature Tf of the gas, we can use the ideal gas law, which states that the pressure times the volume is equal to the number of moles times the gas constant times the temperature (PV = nRT).

Since the pressure and the number of moles are constant, we can write the equation as follows:

T1/V1 = T2/V2

Given that the initial temperature Ti = 1.0°C = 274.15 K (since the Kelvin scale starts at absolute zero, -273.15°C), the initial volume V1 = 1 (let's assume this is 1 unit of volume for simplicity), and the final volume V2 = 9V1 = 9, we can substitute these values into the equation:

274.15 K / 1 = T2 / 9

Solving for T2 gives us:

T2 = 9 * 274.15 K = 2467.35 K

So, the new temperature Tf of the gas is 2467.35 K.

(b) The work done on the gas during the expansion can be calculated using the formula for work done at constant pressure:

W = P * ΔV

where P is the pressure and ΔV is the change in volume. Given that the pressure P = 1.00 atm = 101.325 J/(m^3*K) (converting atmospheres to the SI unit for pressure), the initial volume V1 = 1, and the final volume V2 = 9, we can calculate the change in volume as follows:

ΔV = V2 - V1 = 9 - 1 = 8

Substituting these values into the formula gives us:

W = 101.325 J/(m^3*K) * 8 = 810.6 J

So, the work done on the gas during the expansion is 810.6 J, or 0.8106 kJ (since 1 kJ = 1000 J).