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horrorfan [7]
4 years ago
7

A certain gas is present in a 15.0 LL cylinder at 2.0 atmatm pressure. If the pressure is increased to 4.0 atmatm the volume of

the gas decreases to 7.5 LL . Find the two constants kiki, the initial value of kk, and kfkf, the final value of kk, to verify whether the gas obeys Boyle’s law by entering the numerical values for kiki and kfkf in the space provided.
Chemistry
1 answer:
Helga [31]4 years ago
6 0

Explanation:

According to Boyle's law,  pressure of a gas is inversely proportional to its volume at constant temperature and moles.

Mathematically,       P = \frac{k}{V}

where,     k = proportionality constant

Also, formula for initial pressure and volume is as follows.

              P_{i} = \frac{k_{i}}{V_{i}}

or,           k_{i} = P_{i} \times V_{i}

                          = 2 atm \times 15 L

                          = 30 atm L

Now, we will calculate the value of k_{f} as follows.

             k_{f} = P_{f} \times V_{f}

                         = 4 atm \times 7.5 L

                         = 30 atm L

Hence, as k_{i} = k_{f} this means that it signifies that gas obeys boyle's law.

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The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
hammer [34]

Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

3 0
4 years ago
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3 years ago
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meriva

Answer:

\large \boxed{\text{77.4 mL}}

Explanation:

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