Question

A 25 L cylinder contains 128 g of nitrogen gas at 10oC. How many grams of nitrogen must be added to increase the pressure to 5.00 atm assuming ideal gas behavior

Answer 1

To increase the pressure to 5.00 atm in a 25 L cylinder containing 128 g of nitrogen gas at 10oC, you need to add an additional 2.55 mol of nitrogen gas.

Step-by-step explanation:

To solve this problem, we need to use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature. To find the number of moles of nitrogen gas in the cylinder, we can rearrange the ideal gas law equation to solve for n:

n = PV / RT

Before adding more nitrogen gas, let's calculate the initial number of moles of nitrogen gas in the cylinder:

n = (1 atm) (25 L) / [(0.0821 L * atm / (mol * K)) * (10°C + 273.15)]

n = 1.02 mol

To increase the pressure to 5.00 atm, we need to find the additional number of moles of nitrogen gas needed. The ideal gas law equation can be rearranged to solve for P when n and V are known:

P = nRT / V

Let's calculate the additional number of moles of nitrogen gas needed:

n = (5.00 atm) (25 L) / [(0.0821 L * atm / (mol * K)) * (10°C + 273.15 + 273.15)]

n = 2.55 mol

Therefore, you need to add an additional 2.55 mol of nitrogen gas to increase the pressure to 5.00 atm.

Answer 2

Answer : The amount of nitrogen added must be, 22.64 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

`PV=nRT`

where,

P = Pressure of
`N_2` gas = 5.00 atm

V = Volume of
`N_2` gas = 25 L

n = number of moles
`N_2` = ?

R = Gas constant =
`0.0821L.atm/mol.K`

T = Temperature of
`N_2` gas =
`10^oC=273+10=283K`

Putting values in above equation, we get:

`5.00atm* 25L=n* (0.0821L.atm/mol.K)* 283K`

`n=5.38mol`

Now we have to calculate the mass nitrogen gas.

`\text{Mass of }N_2=\text{Moles of }N_2* \text{Molar mass of }N_2`

Molar mass of nitrogen gas = 28 g/mol

`\text{Mass of }N_2=5.38mol* 28g/mol=150.64g`

Now we have to calculate the amount of nitrogen must be added.

Amount of nitrogen must be added = 150.64 g - 128 g

Amount of nitrogen must be added = 22.64 g

Thus, the amount of nitrogen added must be, 22.64 grams.