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3 years ago

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Warming a gas at constant volume and a pressure of 1.50 atm from 27 degrees Celsius to 127 degrees Celsius gives a final pressur

e of

1 answer:

tankabanditka [31]3 years ago

8 0

Answer:

Final pressure of gas = 2 atm

Explanation:

Given data:

Initial pressure of gas = 1.50 atm

Initial temperature = 27°C

Final temperature = 127°C

Final pressure of gas = ?

Solution:

Initial temperature = 27°C (27+273 = 300 K)

Final temperature = 127°C ( 127+273 = 400 K)

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

1.50 atm / 300 K = P₂/400 K

P₂ = 1.50 atm × 400 K / 300 K

P₂ = 600 atm. K /300 K

P₂ = 2 atm

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An exacuted glass vessel weighs 50 g when empty, 148g when filled with an liquid of density o.989/cc and 50.5 g whenfilled with

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MW of gas : 124.12 g/mol

<h3>Further explanation </h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of density

The unit of density can be expressed in g/cm³ or kg/m³

Density formula:

$\large {\boxed {\bold {\rho ~ = ~ \frac {m} {V}}}}$

ρ = density

m = mass

v = volume

glass vessel wieight = 50 g

glass vessel + liquid = 148 ⇒ liquid = 148 - 50 =98 g

volume of glass vessel :

$\tt V=\dfrac{m}{\rho}=\dfrac{98}{0.989}=99.1~ml$

An ideal gas :

m = 50.5 - 50 = 0.5 g

P = 760 mmHg = 1 atm

T = 300 K

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Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

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The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

<u>Answer:</u> The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

<u>Explanation:</u>

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

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