A monatomic ideal gas (Cv = 3/2 R) undergoes an isothermal expansion at 300 K, as the volume increased from .05 m^3 to .15 m^3 The final pressure is 130kPa The heat transfer to the gas, in kJ, is closest to?

the answer is zero

To find the heat transfer to the gas during an isothermal expansion, we can use the first law of thermodynamics:

Q = ΔU + W

Where Q is the heat transfer, ΔU is the change in internal energy, and W is the work done by the gas.

In an isothermal process, the internal energy remains constant, so ΔU = 0.

Now we can calculate the work done by the gas. The work done in an isothermal expansion is given by:

W = nRT ln(V2/V1)

Where n is the number of moles of gas, R is the ideal gas constant, T is the temperature, and V2/V1 is the ratio of the final to the initial volume.

Given that the gas is monatomic, Cv = 3/2 R. We can use this relationship to find the number of moles of gas:

Cv = (3/2)R
(3/2)R = (3/2)(8.314 J/mol·K) = 12.471 J/mol·K

Since Cv = (3/2)R, Cp = (5/2)R. The relationship between Cp and Cv for monatomic ideal gases is Cp = Cv + R.

So Cp = Cv + R = (3/2)R + R = (5/2)R.

Since Cp = (5/2)R, we can use this relationship to find the number of moles of gas:

Cp = (5/2)R
(5/2)R = (5/2)(8.314 J/mol·K) = 20.785 J/mol·K

Now we can calculate the number of moles of gas using the given equation:

n = CΔT = CpΔT / ∆U

where Cp is the molar heat capacity at constant pressure, ΔT is the change in temperature, and ∆U is the change in internal energy.

ΔT = 300 K - 300 K = 0 K

Since ΔT = 0 K, we find that ∆U = nCv∆T = 0 J

Therefore, n = 0 J / 20.785 J/mol·K = 0 mol

Since n = 0 mol, the number of moles of gas is zero.

Finally, we can calculate the work done by the gas using the formula:

W = nRT ln(V2/V1)

Since n = 0 mol, the work done by the gas is zero as well.

Now we can go back to the first law of thermodynamics:

Q = ΔU + W

Since ΔU = 0 J and W = 0 J, we find that Q = 0 J.

Therefore, the heat transfer to the gas during the isothermal expansion is 0 kJ.