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ehidna [41]
3 years ago
12

A sample of a gas is in a sealed container. The pressure of the gas is 565 torr , and the temperature is 27 ∘C . If the temperat

ure changes to 71 ∘C with no change in volume or amount of gas, what is the new pressure, P2, of the gas inside the container?
Chemistry
1 answer:
Radda [10]3 years ago
5 0

Answer:

P₂ = 647 torr

Explanation:

Given data:

Initial pressure = 565 torr

Initial temperature = 27°C

Final temperature = 71°C

Final pressure = ?

Solution:

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

Final temperature = 71°C (71+273 = 344 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:

565 torr / 300K = P₂/344 K

P₂ = 565 torr × 344 K / 300 K

P₂ = 194360 torr. K /293 K

P₂ = 647 torr

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Answer:

A)  1059 J/mol

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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

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A).

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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\,)

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