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A 280 mL sample of gas exerts 322.5 mm Hg pressure at 26 °C. What pressure does it exert at 69.1 °C if the volume expands to 716.4 mL?

O a. 144
O b. 335
O c. 73.3
O d. 470

1 Answer

2 votes

To solve this problem, we can use the combined gas law, which combines Boyle's law, Charles's law, and Gay-Lussac's law. The combined gas law formula is:

(P1 * V1) / T1 = (P2 * V2) / T2

where:

P1 = Initial pressure (in mm Hg)

V1 = Initial volume (in mL)

T1 = Initial temperature (in Kelvin)

P2 = Final pressure (in mm Hg)

V2 = Final volume (in mL)

T2 = Final temperature (in Kelvin)

Given data:

P1 = 322.5 mm Hg

V1 = 280 mL

T1 = 26 °C = 26 + 273.15 = 299.15 K (converted to Kelvin)

V2 = 716.4 mL

T2 = 69.1 °C = 69.1 + 273.15 = 342.25 K (converted to Kelvin)

Now, let's plug in the values and solve for P2:

(P1 * V1) / T1 = (P2 * V2) / T2

(322.5 * 280) / 299.15 = (P2 * 716.4) / 342.25

Solve for P2:

P2 = (322.5 * 280 * 342.25) / (299.15 * 716.4)

P2 ≈ 144 mm Hg

Therefore, the pressure exerted by the gas at 69.1 °C and a volume of 716.4 mL is approximately 144 mm Hg. The correct answer is option "a. 144".

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