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BabaBlast [244]
3 years ago
7

The average lung capacity of a human is 6.0L.

Chemistry
1 answer:
Darya [45]3 years ago
8 0

Answer:

(a) 0.25 mol

(b) 0.11 mol

(c) 8.77 mol

Explanation:

(a)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 298 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

1.00 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1.00\times 6.0}{0.0821\times 298}=0.25mol

(b)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 0.296 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 200 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

0.296 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 200K\\\\n=\frac{0.296\times 6.0}{0.0821\times 200}=0.11mol

(c)

We use the equation given by ideal gas which follows:

PV=nRT

where,

P = pressure of the gas = 30 atm

V = Volume of the gas = 6.0 L

T = Temperature of the gas = 250 K

R = Gas constant = 0.0821\text{ L.atm }mol^{-1}K^{-1}

n = number of moles = ?

Putting values in above equation, we get:

30 atm\times 6.0L=n\times 0.0821\text{ L.atm }mol^{-1}K^{-1}\times 250K\\\\n=\frac{30\times 6.0}{0.0821\times 250}=8.77mol

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Answer:

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Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

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Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J

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