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3 years ago

A 5.0 L flask contains 0.60 g of 02 at a temperature of 22*C, what is the pressure inside the flask?

Chemistry

1 answer:

Kruka [31]3 years ago

4 0

Answer:

The pressure inside the flask is 0,09 atm

Explanation:

We calculate the number of moles of O2, we convert the Temperature into unit KELVIN and use the formula of the ideal gases to calculate the pressure:

Weight 1 mol of 02= 15,999g x2= 32g

32g---1mol 02

0,60g---x=(0,60gx1mol O2)/32g= 0,019mol O2

T(K)=273+22=295K

PV=nRT

Px5.0l= 0,019mol x 0,082 lxatm/Kxmolx295 K

P=(0,019mol x 0,082 lxatm/Kxmolx295 K)/5.0l

P=0,09atm

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3 years ago

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Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

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If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ( $s_{2} - s_{1}$ ), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$ , $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$ , $c_{w} = 4190\,\frac{J}{kg\cdot K}$ , $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$ , then the change in entropy for the entire process is: