A 1.50 rubber balloon balloon is filled with carbon dioxide gas at a temperature of 0.00 degrees celsius and a pressure at 1.00 atm. The density of carbon dioxide under these conditions are 1.98 g/L.

At 50.0 degrees celsius, the balloon has a volume of 1.78 L. Calculate the carbon dioxide density at this temperature.
Please help!!! And show clears steps, thanks!!

I'm sorry I meant to put a 1.50 L of rubber ballon!

Step 1:

Considering the gas equation for the first condition:

PV = nRT
n = PV/RT
= 1*1.5/0.0821*273
= 1.5/22.41
= 0.066 mol

So we now know that the number of moles of CO2, which will remain fixed, is 0.066 mol.

Step 2:

In the new case -

Total mass of CO2 = Molar mass * number of moles
= 44 * 0.066
= 2.904

Density = Mass/Volume
= 2.904g/1.78L
= 1.63g/L

I agree with the way the problem is worked above; however, I would round the 0.066 mols to 0.067 which makes the new density at 323 K to be 1.66 g/L which to to two significant figures is essentially the same.

Here is much shorter way to work the problem. Since density = g/L and grams is constant, then density varies as T. If T goes up we know volume goers up and a larger volume means a smaller density. So 1.98 g/L x (273/323) = 1.67 g/L.

Thank you very much Arora and DrBob!! You guys were very helpful :)

To calculate the carbon dioxide density at 50.0 degrees Celsius, we need to use the Ideal Gas Law, which states:

PV = nRT

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

First, let's convert the given temperature of 50.0 degrees Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 50.0 + 273.15
T(K) = 323.15 K

Now, we can rearrange the Ideal Gas Law equation to solve for density (D):

D = (molar mass * P) / (R * T)

The molar mass of carbon dioxide (CO2) is 44.01 g/mol.

Step 1: Calculate the number of moles (n) in the balloon at the initial conditions.
Given:
P = 1.00 atm
T = 0.00 degrees Celsius = 273.15 K (converted to Kelvin)

Using the Ideal Gas Law equation, we can solve for n:
n = PV / (RT)
n = (1.00 atm * V) / (0.0821 L·atm/(mol·K) * 273.15 K)
n = V / 22.414

Substituting the given volume (V = 1.78 L):
n = 1.78 L / 22.414
n ≈ 0.08 mol

Step 2: Calculate the mass of carbon dioxide (m) in the balloon at the initial conditions.
The mass (m) is equal to the number of moles (n) multiplied by the molar mass (MM):
m = n * MM
m = 0.08 mol * 44.01 g/mol

Thus, the mass (m) at the initial conditions is approximately 3.52 g.

Step 3: Calculate the density (D) at the initial conditions.
The density (D) is equal to the mass per unit volume:
D = m / V
D = 3.52 g / 1.78 L

Thus, the density (D) at the initial conditions is approximately 1.98 g/L (as given in the question).

Now that we have determined the initial density, let's move on to calculating the density at 50.0 degrees Celsius.

Step 4: Calculate the volume (V) at 50.0 degrees Celsius.
Given:
T = 50.0 degrees Celsius = 323.15 K (as calculated earlier)
V = 1.78 L (as given in the question)

Step 5: Calculate the density (D) at 50.0 degrees Celsius.
Using the Ideal Gas Law equation, we can solve for the number of moles (n) at the new conditions:
n = PV / (RT)
n = (P * V) / (R * T)

Substituting the given values:
n = (1.00 atm * 1.78 L) / (0.0821 L·atm/(mol·K) * 323.15 K)

Step 6: Calculate the mass (m) at 50.0 degrees Celsius.
The mass (m) is equal to the number of moles (n) multiplied by the molar mass (MM):
m = n * MM

Step 7: Calculate the density (D) at 50.0 degrees Celsius.
The density (D) is equal to the mass (m) divided by the volume (V):
D = m / V

By following these steps and assigning the appropriate values at each stage, you should be able to calculate the density of carbon dioxide at 50.0 degrees Celsius.