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3 years ago

5. How can you decrease the pressure of a gas in a container without changing the volume of the gas?

Chemistry

2 answers:

nevsk [136]3 years ago

6 0

Explanation:

Gay-Lussac Law:

This law states that pressure of the gas is directly proportional to the temperature of the gas at constant volume.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$ (at constant volume )

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

So in order to decrease the pressure of the gas in container decrease the temperature of the gas present in the container due to which pressure of the gas will also get decrease as per as Gay-Lussac law.

mars1129 [50]3 years ago

3 0

Answer:

By increaing the size of container

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Radda [10]

Answer: (1). There are 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP.

(2). The density of gaseous arsine is 3.45 g/L.

Explanation:

1). At STP the pressure is 1 atm and temperature is 273.15 K. So, using the ideal gas equation number of moles are calculated as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

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T = temperature

Substitute the values into above formula as follows.

$PV = nRT\\1 atm \times 0.372 L = n \times 0.0821 L atm/mol K \times 273.15 K\\n = 0.0165 mol$

2). As number of moles are also equal to mass of a substance divided by its molar mass.

So, number of moles of Arsine $(AsH_{3})$ (molar mass = 77.95 g/mol) is as follows.

$No. of moles = \frac{mass}{molar mass}\\0.0165 mol = \frac{mass}{77.95 g/mol}\\mass = 1.286 g$

Density is the mass of substance divided by its volume. Hence, density of arsine is calculated as follows.

$Density = \frac{mass}{volume}\\= \frac{1.286 g}{0.372 L}\\= 3.45 g/L$

Thus, we can conclude that 0.0165 moles of gaseous arsine (AsH3) occupy 0.372 L at STP and the density of gaseous arsine is 3.45 g/L.

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