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A sample of an ideal gas has a volume of 3.30 L at 14.80 °C and 1.70 atm. What is the volume of the gas at 21.40 °C and 0.988 atm?

Answer in L please.

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Final answer:

The new volume of the gas at 21.40 °C and 0.988 atm, calculated using the combined gas law, is approximately 5.66 L.

Step-by-step explanation:

To find the new volume of the gas, we can use the combined gas law which relates pressure, volume, and temperature of an ideal gas. The combined gas law is given by (P1 × V1) / T1 = (P2 × V2) / T2, where P is pressure, V is volume, and T is temperature in Kelvin. This equation allows us to solve for the unknown volume (V2) when a gas undergoes a change in pressure and temperature.

First, we convert all temperatures to Kelvin by adding 273.15:

Initial temperature T1 = 14.80 °C + 273.15 = 287.95 K

Final temperature T2 = 21.40 °C + 273.15 = 294.55 K

Next, we rearrange the combined gas law to solve for V2:

V2 = (P1 × V1 × T2) / (P2 × T1)

Finally, we plug in the given values and calculate V2:

V2 = (1.70 atm × 3.30 L × 294.55 K) / (0.988 atm × 287.95 K)

V2 ≈ 5.66 L

Therefore, the volume of the gas at 21.40 °C and 0.988 atm is approximately 5.66 L.

User Kukkuz
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use this

(P_(1) V_(1) )/(T_(1) ) = (P_(2) V_(2) )/(T_(2))

User Ncopiy
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8.6k points