Volume of 0.582kg O2 at 1.0x10^5Pa & 1°C: .3788m^3

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The discussion centers on calculating the volume of 0.582 kg of oxygen gas at a pressure of 1.0 × 10^5 Pa and a temperature of 1°C using the ideal gas law. The initial calculation mistakenly used 532 grams instead of the correct 582 grams, leading to an incorrect number of moles. The correct molecular weight of oxygen is confirmed as approximately 32 g/mol, which is essential for accurate calculations. The proper application of the ideal gas equation with the correct mass should yield the expected volume of 0.3788 m^3. The error highlights the importance of using accurate values in gas law calculations.
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Homework Statement


A mass of 0.582 kg of molecular oxygen is contained in a cylinder at a pressure of 1.0 × 10^5 Pa and a temperature of 1°C. What volume does the gas occupy?


Homework Equations



PV = nRT

The Attempt at a Solution



15.9994 * 2 = 31.9988 amu
532g / 31.9988 amu = 16.62604 mol

K=1 + 273 = 274

(16.62604 x 8.3145 x 274)/100,000
= .3788m^3

This is wrong according to my book...Any ideas why?
 
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The problem statement lists 0.582kg, and you use 532g in your calculation.
 
According to google.com: 0.582 kilograms = 582 grams?
 
Yeah, so why'd you use 532 in your calculation? A typo?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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