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What is the pressure of a can with a volume of 250 mL containing 2.3 g of C3H8 if the can is at 298 K?
Wiki User
∙ 12y ago
You must use the equation PV = nRT, where
P = pressure in ATM
V = volumes in L
n = moles of gas
R = gas constant (.0821 mol*ATM/L*K)
T = Temperature in K
2.3g C3H8 / 44.09 g (molecular weight) = 0.052 moles
250mL = 0.25 L
P(0.25 L) = (0.052 moles)(0.0821 mol*ATM/L*K)(298K)
P = 5.09 ATM
Wiki User
∙ 12y ago
To calculate the pressure of the can, we can use the ideal gas law equation: PV = nRT. Given that the volume V is 250 mL, the temperature T is 298 K, and the amount of substance n (moles) of C3H8 can be calculated using the molar mass of C3H8 and the given mass of 2.3 g. Once we have the number of moles, we can solve for the pressure P.
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