To solve this problem, we will use the ideal gas law equation, which is PV = nRT.

1. First, we need to find the number of moles (n) of carbon dioxide (CO2). The molar mass of CO2 is approximately 44.01 g/mol. So, n = mass/molar mass = 3.00 g / 44.01 g/mol = 0.0681 mol.

2. Next, we need to convert the pressure from kPa to atm because R (the ideal gas constant) is usually given in atm. 1 atm = 101.325 kPa, so P = 127.4 kPa / 101.325 kPa/atm = 1.256 atm.

3. The temperature is given in Kelvin (K), so we don't need to convert it.

4. Now we can plug the values into the ideal gas law equation. We are solving for V (volume), so we rearrange the equation to V = nRT/P.

5. Using R = 0.0821 L·atm/mol·K, we get V = (0.0681 mol)(0.0821 L·atm/mol·K)(309 K) / 1.256 atm = 1.78 L.

So, the volume of 3.00 g of carbon dioxide gas at 309K and 127.4 kPa is approximately 1.78 litres.

Question

To solve this problem, we will use the ideal gas law equation, which is PV = nRT.

1. First, we need to find the number of moles (n) of carbon dioxide (CO2). The molar mass of CO2 is approximately 44.01 g/mol. So, n = mass/molar mass = 3.00 g / 44.01 g/mol = 0.0681 mol.

2. Next, we need to convert the pressure from kPa to atm because R (the ideal gas constant) is usually given in atm. 1 atm = 101.325 kPa, so P = 127.4 kPa / 101.325 kPa/atm = 1.256 atm.

3. The temperature is given in Kelvin (K), so we don't need to convert it.

4. Now we can plug the values into the ideal gas law equation. We are solving for V (volume), so we rearrange the equation to V = nRT/P.

5. Using R = 0.0821 L·atm/mol·K, we get V = (0.0681 mol)(0.0821 L·atm/mol·K)(309 K) / 1.256 atm = 1.78 L.

So, the volume of 3.00 g of carbon dioxide gas at 309K and 127.4 kPa is approximately 1.78 litres.

Knowee AI · Accepted Answer

To solve this problem, we will use the ideal gas law equation, which is PV = nRT.

1. First, we need to find the number of moles (n) of carbon dioxide (CO2). The molar mass of CO2 is approximately 44.01 g/mol. So, n = mass/molar mass = 3.00 g / 44.01 g/mol = 0.0681 mol.

2. Next, we need to convert the pressure from kPa to atm because R (the ideal gas constant) is usually given in atm. 1 atm = 101.325 kPa, so P = 127.4 kPa / 101.325 kPa/atm = 1.256 atm.

3. The temperature is given in Kelvin (K), so we don't need to convert it.

4. Now we can plug the values into the ideal gas law equation. We are solving for V (volume), so we rearrange the equation to V = nRT/P.

5. Using R = 0.0821 L·atm/mol·K, we get V = (0.0681 mol)(0.0821 L·atm/mol·K)(309 K) / 1.256 atm = 1.78 L.

So, the volume of 3.00 g of carbon dioxide gas at 309K and 127.4 kPa is approximately 1.78 litres.

Calculate the volume of 3.00 g of carbon dioxide gas at 309K and 127.4 kPa.Give your answer in litres to three significant figures. Do not include units.

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