The enthalpy of vaporization of benzene liquid is 33.9 kJ/mol at 298 K.

How many liters of benzene gas, measured at 298 K and 95.1 mmHg, are formed when 1.54 kJ of heat is absorbed by benzene liquid at a constant temperature of 298 K?

I think we use the ideal gas law, but I'm not sure what to do with the heat.

33.9 kJ/mol x ? mol = 1.54 kJ

Solve for ?mol benzene, then substitute into PV = nRT and solve for volume in L.

To solve this problem, we can use the combined gas law and the concept of heat transfer.

The ideal gas law is given by:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin

To determine the volume of benzene gas formed, we need to consider the heat absorbed by the liquid. Enthalpy is defined as the heat transfer at constant pressure.

The enthalpy change can be calculated using the formula:

ΔH = nΔHvap

Where:
ΔH = change in enthalpy (in kJ)
n = moles of substance
ΔHvap = enthalpy of vaporization (in kJ/mol)

We can rearrange this equation to solve for the number of moles:

n = ΔH/ΔHvap

Then, we can use the ideal gas law to determine the volume of benzene gas:

V = (nRT)/P

Now, let's plug in the given values:

ΔHvap = 33.9 kJ/mol
ΔH = 1.54 kJ
T = 298 K
P = 95.1 mmHg (convert to atm by dividing by 760 mmHg/atm)

First, let's calculate the number of moles of benzene using the enthalpy change equation:

n = (1.54 kJ) / (33.9 kJ/mol)
n ≈ 0.045 mol

Now, let's calculate the volume of benzene gas using the ideal gas law:

V = (0.045 mol) * (0.0821 L·atm/(mol·K)) * (298 K) / (95.1 mmHg / 760 mmHg/atm)
V ≈ 1.15 L

Therefore, approximately 1.15 liters of benzene gas are formed when 1.54 kJ of heat is absorbed by benzene liquid at a constant temperature of 298 K and a pressure of 95.1 mmHg.

To solve this problem, you can use the ideal gas law equation, but you also need to consider the heat absorbed by the liquid. Here's how you can approach it step by step:

1. Convert the enthalpy of vaporization from kJ/mol to J/mol:
Enthalpy of vaporization = 33.9 kJ/mol = 33.9 × 10^3 J/mol

2. Use the equation q = nΔHvap to calculate the number of moles of benzene vaporized:
q = 1.54 kJ = 1.54 × 10^3 J
ΔHvap = 33.9 × 10^3 J/mol

Rearrange the equation to solve for n (number of moles):
n = q / ΔHvap

3. Calculate the number of moles of benzene vaporized:
n = (1.54 × 10^3 J) / (33.9 × 10^3 J/mol)

4. Use the ideal gas law, PV = nRT, to calculate the volume of the benzene gas:
P = 95.1 mmHg = 95.1/760 atm (since 1 atm = 760 mmHg)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 298 K (temperature)
n = (1.54 × 10^3 J) / (33.9 × 10^3 J/mol)

Rearrange the equation to solve for V (volume):
V = (nRT) / P

5. Calculate the volume of the benzene gas:
V = [(1.54 × 10^3 J) / (33.9 × 10^3 J/mol)] × (0.0821 L·atm/(mol·K)) × 298 K / (95.1/760) atm

6. Calculate the final volume by dividing the volume by 1000 (to convert from mL to L) since the units in the question are in liters.

By following these steps, you can calculate the volume of benzene gas formed.