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Reika [66]
3 years ago
9

13. An aerosol spray can of deodorant with a volume of 0.410 L contains 3.0 g of propane gas (CH3) as propellant. What is the pr

essure in the can at 20°C? VRT 3. og festa​
Chemistry
1 answer:
qaws [65]3 years ago
8 0

Answer: The pressure in the can is 4.0 atm

Explanation:

According to ideal gas equation:

PV=nRT

P = pressure of gas = ?

V = Volume of gas = 0.410 L

n = number of moles = \frac{\text {given mass}}{\text {Molar mass}}=\frac{3.0g}{44.1g/mol}=0.068mol

R = gas constant =0.0821Latm/Kmol

T =temperature =20^0C=(20+273)K=293K

P=\frac{nRT}{V}

P=\frac0.068mol\times 0.0820 L atm/K mol\times 293K}{0.410L}=4.0atm

Thus the pressure in the can is 4.0 atm

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If you are given an ideal gas with pressure (p)259,392.00 pa and temperature (T)=200°c of 1 mole Argon gas in a volume 8.8dm3,ca
GuDViN [60]

Answer: R=4.82436 \frac{Pa. m^{3}}{mol. K}

Explanation:

The Ideal Gas equation is:  

P.V=n.R.T  (1)

Where:  

P is the pressure of the gas  

n the number of moles of gas  

R=8.3144598 \frac{Pa. m^{3}}{mol. K} is the gas constant  

T is the absolute temperature of the gas in Kelvin.

V is the volume

It is important to note that the behavior of a real gas is far from that of an ideal gas, taking into account that <u>an ideal gas is a single hypothetical gas</u>. However, under specific conditions of standard temperature and pressure (T=0\°C=273.15 K and P=1 atm=101,3 kPa) one mole of real gas (especially in noble gases such as Argon) will behave like an ideal gas and the constant R will be 8.3144598 \frac{Pa. m^{3}}{mol. K}.

However, in this case we are not working with standard temperature and pressure, therefore, even if we are working with Argon, the value of R will be far from the constant of the ideal gases.

Having this clarified, let's isolate R from (1):

R=\frac{PV}{nT}  (2)

Where:

P=259392 Pa

n=1 mole

T=200\°C=473.15 K is the absolute temperature of the gas in Kelvin.

V=8.8 dm^{3}=0.0088 m^{3}

R=\frac{(259392 Pa)(0.0088 m^{3})}{(1 mole)(473.15 K)}  (3)

Finally:

R=4.82436 \frac{Pa. m^{3}}{mol. K}  

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