BP -164°C ΔHvap 8.136 kJ/mol sg 2.205 J/g°C

MP -182°C ΔHfus 0.928 kJ/mol sl 2.41 J/g°C

Using the data above for methane, CH4, calculate the heat needed to convert 150. g of liquid methane at -175°C to a gas at 25°C.

To calculate the heat needed to convert 150 g of liquid methane at -175°C to a gas at 25°C, we can break the process into two steps: heating the liquid methane from -175°C to its boiling point, and then vaporizing it from its boiling point to the desired temperature.

Step 1: Heating the liquid methane from -175°C to its boiling point.
Given:
- BP (Boiling Point) of methane = -164°C
- ΔHvap (Heat of Vaporization) of methane = 8.136 kJ/mol
- sg (Specific Heat Capacity) of methane = 2.205 J/g°C

First, we need to determine the amount of heat required to heat the liquid methane from -175°C to its boiling point (-164°C).

ΔT = final temperature - initial temperature = -164°C - (-175°C) = 11°C

The heat energy required for this step is given by the formula:

q1 = mass (g) × sg (J/g°C) × ΔT (°C)

Calculating q1:

q1 = 150 g × 2.205 J/g°C × 11°C
q1 = 3638.025 J

Step 2: Vaporizing the liquid methane from its boiling point to the desired temperature.
Given:
- ΔHfus (Heat of Fusion) of methane = 0.928 kJ/mol
- sl (Specific Heat Capacity) of methane (in solid phase) = 2.41 J/g°C

Since we are converting liquid methane to gas, we need to use the heat of vaporization (ΔHvap) instead of the heat of fusion (ΔHfus).

The heat energy required for this step can be calculated using the formula:

q2 = mass (g) × ΔHvap (kJ/mol)

However, we need to convert the mass of methane from grams to moles using its molar mass.

Molar mass of methane (CH4):
(1 × 12.01 g/mol) + (4 × 1.01 g/mol) = 16.05 g/mol

Number of moles of methane in 150 g:
moles = mass (g) / molar mass (g/mol)
moles = 150 g / 16.05 g/mol
moles ≈ 9.34 mol

Calculating q2:

q2 = moles × ΔHvap (kJ/mol)
q2 = 9.34 mol × 8.136 kJ/mol
q2 ≈ 75.78 kJ

Total energy required:
Total heat energy = q1 + q2
Total heat energy ≈ 3638.025 J + 75.78 kJ

Converting 75.78 kJ to Joules:
1 kJ = 1000 J
75.78 kJ = 75.78 × 1000 J ≈ 75780 J

Adding the values:

Total heat energy ≈ 3638.025 J + 75780 J
Total heat energy ≈ 79418.025 J

Therefore, the heat needed to convert 150 g of liquid methane at -175°C to a gas at 25°C is approximately 79418.025 J.

To calculate the heat needed to convert liquid methane to a gas, we need to consider two steps: first, we need to heat the liquid methane from -175°C to its boiling point of -164°C, and then we need to vaporize it at its boiling point.

Step 1: Heating the liquid methane from -175°C to -164°C:
To calculate the heat needed for this step, we can use the heat capacity equation:
q = m * C * ΔT
where:
q is the heat needed
m is the mass of the substance
C is the specific heat capacity
ΔT is the change in temperature

Given data:
m = 150. g (mass of liquid methane)
C = 2.41 J/g°C (specific heat capacity of solid/liquid methane)
ΔT = -164°C - (-175°C) = 11°C

q = 150. g * 2.41 J/g°C * 11°C
q = 4081.5 J

Step 2: Vaporizing the liquid methane at its boiling point (-164°C to gas at 25°C):
To calculate the heat needed for this step, we can use the heat of vaporization equation:
q = n * ΔHvap
where:
q is the heat needed
n is the number of moles of the substance
ΔHvap is the heat of vaporization

To find the number of moles (n), we can use the formula:
n = m / M
where:
m is the mass of the substance
M is the molar mass

Given data:
m = 150. g (mass of liquid methane)

To find the molar mass (M) of methane (CH4), we can use the periodic table:
C (carbon) = 12 g/mol
H (hydrogen) = 1 g/mol

M = 12 g/mol + (4 * 1 g/mol) = 16 g/mol

n = 150. g / 16 g/mol
n = 9.375 mol

Now, we can calculate the heat needed for vaporization:
q = 9.375 mol * 8.136 kJ/mol
q = 76.26 kJ (since 1 kJ = 1000 J, this can also be written as 76,260 J)

Total heat needed = heat for step 1 + heat for step 2
Total heat needed = 4081.5 J + 76260 J
Total heat needed = 80341.5 J

Therefore, the heat needed to convert 150. g of liquid methane at -175°C to a gas at 25°C is 80,341.5 J or 80.34 kJ.