Molar mass CH4 = 16.0 g/mol
* number of moles:
932.3 / 16 => 58.26875 moles
T = 136.2 K
V = 0.560 L
P = ?
R = 0.082
Use the clapeyron equation:
P x V = n x R x T
P x 0.560 = 58.26875 x 0.082 x 136.2
P x 0.560 = 650.76
P = 650.76 / 0.560
P = 1162.07 atm
Answer:
0.962 atm.
97.4 kPa.
731 torr.
14.1 psi.
97,434.6 Pa.
Explanation:
Hello.
In this case, given the available factors equaling 1 atm of pressure, each required pressure turns out:
- Atmospheres: 1 atm = 760 mmHg:

- Kilopascals:: 101.3 kPa = 760 mmHg:

- Torrs: 760 torr = 760 mmHg:

- Pounds per square inch: 14.69 psi = 760 mmHg:

- Pascals: 101300 Pa = 760 mmHg:

Best regards.
Answer:
π = 14.824 atm
Explanation:
wt % = ( w NaCL / w sea water ) * 100 = 3.5 %
assuming w sea water = 100 g = 0.1 Kg
⇒ w NaCl = 3.5 g
osmotic pressure ( π ):
∴ T = 20 °C + 273 = 293 K
∴ C ≡ mol/L
∴ density sea water = 1.03 Kg/L....from literature
⇒ volume sea water = 0.1 Kg * ( L / 1.03 Kg ) = 0.097 L sln
⇒ mol NaCl = 3.5 g NaCL * ( mol NaCL / 58.44 g ) = 0.06 mol
⇒ C NaCl = 0.06 mol / 0.097 L = 0.617 M
⇒ π = 0.617 mol/L * 0.082 atm L / K mol * 293 K
⇒ π = 14.824 atm
Answer:
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