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a 2.00 l cylinder contains 4.28 g of methane, ch₄ , at 3800 mmhg. what is the temperature of the gas?

User GGcupie
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Final answer:

To calculate the temperature of the methane gas, we converted the pressure to atm, determined the number of moles, and used the Ideal Gas Law. The resulting temperature was found to be 367.55 K.

Step-by-step explanation:

To find the temperature of the methane gas in the cylinder, we can use the Ideal Gas Law, which is expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is the temperature in Kelvin.

First, we must convert the given pressure from mmHg to atm because the gas constant R is typically given in liters·atm/mol·K. 1 atm = 760 mmHg, therefore 3800 mmHg = 3800 mmHg ÷ 760 mmHg/atm = 5 atm.

Next, we calculate the number of moles of methane using its molecular weight (16.04 g/mol):
n = mass of CH4 ÷ molar mass of CH4
n = 4.28 g ÷ 16.04 g/mol = 0.267 moles of CH4

Now, we plug in the values into the Ideal Gas Law and solve for T:
PV = nRT
5 atm × 2.00 L = 0.267 moles × (0.0821 liters·atm/mol·K) × T
T = (5 atm × 2.00 L) ÷ (0.267 moles × 0.0821 liters·atm/mol·K)
T = 367.55 K

Therefore, the temperature of the gas is 367.55 K.

User Letholdrus
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