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(a) Two moles of helium gas initially at 257 K and 0.42 atm are compressed isothermally to 1.88 atm. Find the final volume of the gas. Assume that helium behaves as an ideal gas. The universal gas constant is 8.31451 J/K · mol. Answer in units of m3.

b) Find the work done by the gas. Answer in units of kJ.

(c) Find the thermal energy transferred. Answer in units of kJ.

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Answer: (a) The final volume of the gas is 0.078 m^3.

(b) The work done by the gas is -2.2997 kJ.

(c) The thermal energy transferred is 2.2997 kJ.

Explanation: (a) The process is isothermal, which means the temperature remains constant during the compression. Therefore, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

Since the process is isothermal, T is constant, and we can write:

P1V1 = P2V2

where subscripts 1 and 2 refer to the initial and final states, respectively.

We are given that n = 2 mol, P1 = 0.42 atm, P2 = 1.88 atm, and T = 257 K. Therefore, we can solve for V2:

V2 = V1 * P1/P2 = (nRT)/P2 * P1

Substituting the values, we get:

V2 = (2 mol * 8.31451 J/K·mol * 257 K) / (1.88 atm) * (0.42 atm) = 0.078 m^3

Therefore, the final volume of the gas is 0.078 m^3.

(b) The work done by the gas during an isothermal process is given by:

W = -nRT ln(P2/P1)

Substituting the values, we get:

W = -(2 mol) * (8.31451 J/K·mol) * (257 K) * ln(1.88/0.42) = -2299.7 J

Therefore, the work done by the gas is -2299.7 J or -2.2997 kJ (to three significant figures).

(c) Since the process is isothermal, the thermal energy transferred is equal to the work done by the gas:

Q = -W = 2.2997 kJ

Therefore, the thermal energy transferred is 2.2997 kJ.

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