One mole of an ideal gas is sealed in a 22.4-L container at a pressure of 1 atm and a temperature of 273 K. The temperature is t

Question

3 years ago

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One mole of an ideal gas is sealed in a 22.4-L container at a pressure of 1 atm and a temperature of 273 K. The temperature is t

hen increased to 304 K , but the container does not expand. What will the new pressure be?
Part A

The most appropriate formula for solving this problem includes only which variables?

Enter the required variables, separated by commas (e.g., P,V,T).

Q2)A sample of nitrogen gas in a 1.69-L container exerts a pressure of 1.37 atm at 17 ∘C.

-What is the pressure if the volume of the container is maintained constant and the temperature is raised to 327 ∘C?

Q3)A gas mixture with a total pressure of 770 mmHgcontains each of the following gases at the indicated partial pressures: 120 mmHg CO2, 227mmHg Ar, and 190 mmHg O2. The mixture also contains helium gas

-What mass of helium gas is present in a 14.0-L sample of this mixture at 282 K ?

Q4)

A

Calculate the density of oxygen, O2, under each of the following conditions:

STP

1.00 atm and 35.0 ∘C

Express your answers numerically in grams per liter. Enter the density at STP first and separate your answers by a comma.

B

To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 4.1-L bulb, then filled it with the gas at 2.00 atm and 24.0 ∘C and weighed it again. The difference in mass was 9.5 g . Identify the gas.

Express your answer as a chemical formula.

Chemistry

1 answer:

Ket [755] · Answer 1 · 2022-01-31T16:18:54+03:00

Answer:

1.11 atm

(P, T)

2.83 atm

0.740 g

0.179 g/L, 0.158 g/L

N₂

Explanation:

One mole of an ideal gas is sealed in a 22.4-L container at a pressure of 1 atm and a temperature of 273 K. The temperature is then increased to 304 K, but the container does not expand. What will the new pressure be?

Assuming ideal behavior, we can calculate the new pressure (P₂) using Gay-Lussac's law.

$\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}} \\P_{2}=\frac{P_{1}}{T_{1}}.T_{2}=\frac{1atm}{273K} .304K=1.11atm$

The most appropriate formula for solving this problem includes only which variables?

Gay-Lussac's law includes pressure (P) and absolute temperature (T).

Q2) A sample of nitrogen gas in a 1.69-L container exerts a pressure of 1.37 atm at 17 °C. What is the pressure if the volume of the container is maintained constant and the temperature is raised to 327 °C?

Initially the system is at 17°C (290 K) and the temperature is raised to 327°C (600 K). We can calculate the new pressure using Gay-Lussac's law.

$\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}} \\P_{2}=\frac{P_{1}}{T_{1}}.T_{2}=\frac{1.37atm}{290K} .600K=2.83atm$

Q3) A gas mixture with a total pressure of 770 mmHg contains each of the following gases at the indicated partial pressures: 120 mmHg CO₂, 227mmHg Ar, and 190 mmHg O₂. The mixture also contains helium gas .

What mass of helium gas is present in a 14.0-L sample of this mixture at 282 K?

First, we have to calculate the pressure of Helium. We know that the total pressure is the sum of partial pressures.

Ptotal = pCO₂ + pAr + pO₂ + pHe

pHe = Ptotal - pCO₂ - pAr - pO₂

pHe = 770mmHg - 120mmHg - 227mmHg - 190mmHg=233mmHg

We can calculate the moles of Helium using the ideal gas equation.

$P.V=n.R.T\\n=\frac{P.V}{R.T} =\frac{233mmHg.14.0L}{(0.08206atm.L/mol.K).282K} .\frac{1atm}{760mmHg} =0.185mol$

The molar mass of He is 4.00g/mol.

$0.185mol.\frac{4.00g}{mol} =0.740g$

Calculate the density of oxygen, O₂, under each of the following conditions:

STP

1.00 atm and 35.0 ∘C

Express your answers numerically in grams per liter. Enter the density at STP first and separate your answers by a comma.

STP stands for Standard Temperature and Pressure. The standard temperature is 273 K and the standard pressure is 1 atm.

We can calculate the density using the following expression:

$\rho=\frac{P.M}{R.T} =\frac{1.00atm.4.00g/mol}{(0.08206atm.L/mol.K).273K} =0.179 g/L$

At 1.00 atm and 35.0 °C (308 K)

$\rho=\frac{P.M}{R.T} =\frac{1.00atm.4.00g/mol}{(0.08206atm.L/mol.K).308K} =0.158 g/L$

To identify a diatomic gas (X₂), a researcher carried out the following experiment: She weighed an empty 4.1-L bulb, then filled it with the gas at 2.00 atm and 24.0 ∘C and weighed it again. The difference in mass was 9.5 g . Identify the gas. Express your answer as a chemical formula.

We will look for the molar mass of the compound using the ideal gas equation.

$P.V=n.R.T=\frac{m}{M} .R.T\\M=\frac{m.R.T}{P.V} =\frac{9.5g \times (0.08206atm.L/mol.K)\times 297K }{2.00atm \times 4.1L} =28g/mol$