Question

1) a sample of gas at 2.50 atm pressure and 12 degrees celsius has a volume of 35 mL. the pressure is lowered to 0.85 atm and the temperature increases to 298K. what is the final volume of the gas in L? 2) find the moles of gas present at a temperature of 45 degrees celsius, volume of 67.5 L and a pressure of 637 torr 3) a gas is initially at a pressure of 7.5 atm and a volume of 2.7 L. when the pressure is increased to 9.5 atm, what is the bew volume?

Answer 1

For the given questions: 1) The final volume of the gas is 0.108 L. 2) The moles of gas present are 2.24 mol. 3) The new volume of the gas is 2.14 L.

Step-by-step explanation:

1) To calculate the final volume of the gas, we can use the combined gas law formula:

P1V1/T1 = P2V2/T2

Using the given values:

P1 = 2.50 atm, V1 = 35 mL = 0.035 L, T1 = 12°C + 273 = 285K

P2 = 0.85 atm, T2 = 298K

Plugging in the values and solving for V2:

(2.50 atm)(0.035 L)/(285K) = (0.85 atm)(V2)/(298K)

Simplifying the equation gives:

V2 = (2.50 atm)(0.035 L)(298K)/(0.85 atm)(285K)

V2 = 0.108 L

Therefore, the final volume of the gas is 0.108 L.

2) To find the moles of gas, we can use the ideal gas law formula:

PV = nRT

Using the given values and converting the temperature to Kelvin:

P = 637 torr = 0.837 atm, V = 67.5 L, T = 45°C + 273 = 318K, R = 0.0821 L.atm/mol.K

Plugging in the values and solving for n:

(0.837 atm)(67.5 L) = n(0.0821 L.atm/mol.K)(318K)

Simplifying the equation gives:

n = (0.837 atm)(67.5 L)/(0.0821 L.atm/mol.K)(318K)

n = 2.24 mol

Therefore, the moles of gas present are 2.24 mol.

3) To find the new volume, we can use Boyle's Law:

P1V1 = P2V2

Using the given values:

P1 = 7.5 atm, V1 = 2.7 L, P2 = 9.5 atm

Plugging in the values and solving for V2:

(7.5 atm)(2.7 L) = (9.5 atm)(V2)

Simplifying the equation gives:

V2 = (7.5 atm)(2.7 L)/(9.5 atm)

V2 = 2.14 L

Therefore, the new volume of the gas is 2.14 L.

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