2 moles of ideal gas at 300K, initial volume 25 L, are compressed under an external pressure at 3600 torr. Assuming the temp and the internal energy of gas do not change, what is the entropy change of the gas? (Assume the heat flow is reversible)

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To find the entropy change (ΔS) of the gas, we need to use the equation:

ΔS = nRln(V2/V1)

Where:
ΔS is the entropy change
n is the number of moles of gas
R is the ideal gas constant (8.314 J/(mol·K))
ln is the natural logarithm
V2 is the final volume of the gas
V1 is the initial volume of the gas

In this case, we know that n = 2 moles, V1 = 25 L, and V2 is the final volume of the gas.

To find V2, we can use the ideal gas law:

PV = nRT

Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature in Kelvin

In this case, the initial pressure (P1) is 3600 torr, the final volume (V2) is unknown, the number of moles (n) is 2, the ideal gas constant (R) is 0.08206 L·atm/(mol·K), and the temperature (T) is 300 K.

First, let's convert the initial pressure from torr to atm:
3600 torr * (1 atm / 760 torr) = 4.74 atm

Now, we can rearrange the ideal gas law equation to solve for V2:
V2 = (P2 * V1) / P1

Plugging in the values:
V2 = (4.74 atm * 25 L) / 3600 torr

Now, we need to convert the final volume from liters to m^3 (SI units):
V2 = ((4.74 atm * 25 L) / 3600 torr) * (1.01325 × 10^5 Pa / 1 atm) * (1 m^3 / 1000 L)

Calculating this, we find V2 ≈ 0.03465 m^3.

Finally, we can substitute the values of n, R, and V2 into the entropy change equation to find ΔS:
ΔS = (2 moles * 8.314 J/(mol·K)) * ln(0.03465 m^3 / 25 L)

Simplifying this expression will give us the entropy change of the gas (ΔS).