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3 years ago

Which measure of a gas does the expression nRT represent? (1 point)

1 answer:

7 0

The Ideal Gas Equation
The term pVnRT is also called the compression factor and is a measure of the ideality of the gas. An ideal gas will always equal 1 when plugged into this equation……

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ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.

timofeeve [1]

Complete question:

ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.

Answer:

The magnitude of q for the process 568 J.

Explanation:

Given;

change in internal energy of the gas, ΔU = 475 J

work done by the gas, w = 93 J

heat added to the system, = q

During gas expansion process, heat is added to the gas.

Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.

ΔU = q - w

q = ΔU + w

q = 475 J + 93 J

q = 568 J

Therefore, the magnitude of q for the process 568 J.

6 0

3 years ago

The heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its temp

LekaFEV [45]

Answer:

$Q=9.2x10^5J$

$t=614s=10.2min$

Explanation:

Hello,

In this case, we can compute the energy by using the following formula for air:

$Q=nCp\Delta T$

Whereas the moles of air are computed via the ideal gas equation at room temperature inside the 5.5m x 6.5m x 3.0m-room:

$n=\frac{PV}{RT}\\\\V=5.5m*6.5m*3.0m=107.25m^3*\frac{1000L}{1m^3}=107250L\\ \\n=\frac{1atm*107250L}{0.082\frac{atm*L}{mol*K}*298.15K}\\ \\n=4386.8mol$

Now, we are able to compute heat, by considering that the temperature raise is given in degree Celsius or Kelvins as well:

$Q=4386.8mol*21\frac{K}{mol*K}*10K \\\\Q=9.2x10^5J$

Finally, we compute the time required for the heating by considering the heating rate and the required heat, shown below:

$t=\frac{9.2x10^5J}{1.5\frac{kJ}{s}*\frac{1000J}{1kJ} } \\\\t=614s=10.2min$

Regards.

8 0

3 years ago

Two linear hydrocarbons, Hexane (C6H14) and Heptane (C7H16), form pretty much an ideal solution at any composition. A solution i

Anettt [7]

Answer:

$y_{C_6H_{14}}=0.92$

Explanation:

Hello,

At first, we compute liquid-phase molar fractions:

$n_{C_6H_{14}}=463.96 g*\frac{1mol}{86g} =5.3949molC_6H_{14}\\n_{C_7H_{16}}=667.71 g*\frac{1mol}{100g} =6.6771molC_7H_{16}\\x_{C_6H_{14}}=\frac{5.3949}{5.3949+6.6771} =0.447\\x_{C_7H_{16}}=1-x_{C_6H_{14}}=0.553$

Now, by means of the fugacity concept, for hexane, for instance, we have:

$f_{C_6H_{14}}^V=f_{C_6H_{14}}^L\\y_{C_6H_{14}}p_T=x_{C_6H_{14}}p_{C_6H_{14}}$

In this manner, at 25 °C the vapor pressure of hexane and heptane are 0.198946 atm and 0.013912 atm repectively, thus, the total pressure is:

$p_T=x_{C_6H_{14}}p_{C_6H_{14}}+x_{C_7H_{16}}p_{C_7H_{16}}\\p_T=0.447*0.198946 atm +0.553*0.013912 atm=0.096622atm$

Finally, from the hexane's fugacity equation, we find its mole fraction in the vapour as:

$y_{C_6H_{14}}=\frac{x_{C_6H_{14}}p_{C_6H_{14}}}{p_T}=\frac{0.447*0.198946 atm}{0.096622atm} \\y_{C_6H_{14}}=0.92$

Best regards.

3 0

3 years ago

Read 2 more answers

Which of the following diatomic molecules DO NOT form single bonds?

Karolina [17]

Answer:

Br2?? I think I tried looking it up on quizlet but it didnt work

8 0

4 years ago

What are the possible values of n and ml for an electron in a 5d orbital?

liberstina [14]

Answer:

The answer to your question is n = 5, l = 2, m can be -2, -1, 0, 1 or 2

Explanation:

Data

orbital = 5d

values of n, l, m

Process

1.- Determine the value of n

n is the coefficient of the orbital, in this problem n = 5

2.- Determine the value of l

l takes values depending in the sublevel of energy,

if the sublevel is s then l = 0

p l = 1

d l = 2

f l = 3

For this problem l = 2

3.- Determine the value of m

when l = 2, m takes values of -2, - 1, 0, 1 or 2

3 0

3 years ago