26 g of methane CH4 has a pressure of 450 kPa at 250 degrees celcius find the volume occupied by the gas

PV =nRT and n = grams/molar mass

I would change kPa to atm. T must be in kelvin. Note the correct spelling of celsius.

Your gonna be looking at 15 L right there

To find the volume occupied by the gas, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure (in kPa)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K) or 8.314 J/(mol·K))
T = temperature (in Kelvin)

First, we need to convert the given temperature from degrees Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 250 + 273.15
T(K) = 523.15 K

Next, we need to calculate the number of moles of gas:
n = mass / molar mass

The molar mass of methane (CH4) is:
12.01 g/mol (C) + 4(1.01 g/mol) = 16.04 g/mol

n = 26 g / 16.04 g/mol
n = 1.62 mol

Now, we can substitute the values into the ideal gas law equation:

PV = nRT

V = (nRT) / P
V = (1.62 mol)(0.0821 L·atm/(mol·K))(523.15 K) / 450 kPa

Since the ideal gas constant is given in L·atm/(mol·K) and the pressure unit is in kPa, we need to convert the pressure to atm:

1 atm = 101.325 kPa

V = (1.62 mol)(0.0821 L·atm/(mol·K))(523.15 K) / (450 kPa * (1 atm / 101.325 kPa))
V = 0.4297 L

Therefore, the volume occupied by the gas is approximately 0.4297 liters.

To find the volume occupied by the gas, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure of the gas in kilopascals (kPa)
V = volume of the gas in liters (L)
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/(K·mol) or 8.31 J/(K·mol))
T = temperature of the gas in Kelvin (K)

First, we need to convert the temperature from Celsius to Kelvin. The Celsius temperature can be converted to Kelvin by adding 273.15:

T(K) = T(°C) + 273.15

In this case:
T(K) = 250°C + 273.15 = 523.15 K

Next, we need to calculate the number of moles. We can use the formula:

n = mass / molar mass

The molar mass of methane (CH4) is:
molar mass(CH4) = (1 mol C) + (4 mol H) = 12.01 g/mol + (4 * 1.01 g/mol) = 16.05 g/mol

In this case:
n = 26 g / 16.05 g/mol ≈ 1.62 mol

Now, we can plug the values into the ideal gas law equation and solve for V:

PV = nRT
V = nRT / P

V = (1.62 mol)(0.0821 L·atm/(K·mol))(523.15 K) / 450 kPa

To make the units match, we need to convert kilopascals to atmospheres:
1 atm = 101.3 kPa

V ≈ (1.62 mol)(0.0821 L·atm/(K·mol))(523.15 K) / (450 kPa / 101.3 kPa/atm)
V ≈ (1.62 mol)(0.0821 L·atm/(K·mol))(523.15 K)(101.3 kPa) / 450 kPa

Now we can calculate the volume using the above expression.