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3 years ago

An ideal gas in a sealed container has an initial volume of 2.80 L. At constant pressure, it is cooled to 18.00 °C, where its

Chemistry

1 answer:

Aleks04 [339]3 years ago

6 0

Answer:

$T_1=-91.18\°C$

Explanation:

Hello there!

In this case, given the T-V variation, we understand it is possible to apply the Charles' law as shown below:

$\frac{T_1}{V_1}= \frac{T_2}{V_2}$

Thus, since we are interested in the initial temperature, we can solve for T1, plug in the volumes and use T2 in kelvins:

$T_1= \frac{T_2V_1}{V_2}\\\\T_1= \frac{(18.00+273.15)K(1.75L)}{(2.80L)}\\\\T_1=182K-273.15\\\\T_1=-91.18\°C$

Best regards!

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Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

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Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

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Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

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a rectangular piece of aluminum foil measures 13.72 cm x 8.63 cm and has a mass of 3.1 g. Find how thick it is

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The two dimensions of aluminum foil are given 13.72 cm and 8.63 cm respectively with mass 3.1 g.

The density of aluminum is $2.7 g/cm^{3}$ . It is defined as mass per unit volume thus, volume of aluminum can be calculated as follows:

$V=\frac{m}{d}$

Putting the values.

$V=\frac{3.1 g}{2.7 g/cm^{3}}=1.148 cm^{3}$

The volume of cuboid is $l\times b\times h$ , the length and breadth are given, height can be calculated as follows;

$h=\frac{V}{l\times b}$

Putting the values,

$h=\frac{1.148 cm^{3}}{(13.72 cm)(8.63 cm)}=0.009695 cm$

Or, approximately 9.7\times 10^{-3}cm

Therefore, thickness of aluminum foil is $9.7\times 10^{-3}cm$

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