The pressure in a vessel that contained pure oxygen dropped from 2000torr to 1500 torr in 55 min as oxygen leaked.when some vessel was filled with another gas the pressure dropped from 2000torr to 1500torr in 85 min.find the molar mass of another gas

To determine the molar mass of the other gas, we can use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. This principle allows us to compare the rates at which two different gases leak from a vessel under similar conditions.

Understanding the Problem

We have two scenarios here:

• Pure oxygen leaking from a vessel, where the pressure drops from 2000 torr to 1500 torr in 55 minutes.
• Another gas leaking from a vessel, where the pressure also drops from 2000 torr to 1500 torr, but this time in 85 minutes.

Calculating the Rate of Effusion

First, we need to calculate the rate of pressure drop for both gases. The rate of effusion can be expressed as the change in pressure over time.

For Oxygen:

The change in pressure for oxygen is:

• Initial pressure = 2000 torr
• Final pressure = 1500 torr
• Change in pressure = 2000 torr - 1500 torr = 500 torr

The rate of effusion for oxygen (R_O) is:

R_O = Change in pressure / Time = 500 torr / 55 min = 9.09 torr/min

For the Other Gas:

Similarly, for the other gas, the change in pressure is the same:

• Change in pressure = 500 torr

The rate of effusion for the other gas (R_G) is:

R_G = Change in pressure / Time = 500 torr / 85 min = 5.88 torr/min

Applying Graham's Law

According to Graham's law, we can relate the rates of effusion of the two gases to their molar masses:

R_O / R_G = √(M_G / M_O)

Where:

• M_O = Molar mass of oxygen (approximately 32 g/mol)
• M_G = Molar mass of the other gas (unknown)

Substituting the Values

Now we can substitute the rates we calculated:

9.09 / 5.88 = √(M_G / 32)

Squaring Both Sides

To eliminate the square root, we square both sides:

(9.09 / 5.88)² = M_G / 32

Calculating the left side:

(9.09 / 5.88)² ≈ 2.67

Solving for Molar Mass

Now we can solve for M_G:

M_G = 32 * 2.67 ≈ 85.44 g/mol

Final Result

The molar mass of the other gas is approximately 85.44 g/mol. This value suggests that the gas could be something like propane (C₃H₈), which has a molar mass of about 44 g/mol, or possibly another heavier gas. Understanding the context of the gases involved can help narrow down the possibilities further.