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Viefleur [7K]
3 years ago
14

3.00 L of a gas is collected at 35.0°C and 705.0 mmHg. What is the volume at STP?

Chemistry
1 answer:
makvit [3.9K]3 years ago
8 0

Answer:

2.47L

Explanation:

Using the combined gas law equation as follows:

P1V1/T1= P2V2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question;

P1 = 705mmHg

P2 = 760mmHg (STP)

V1 = 3.00L

V2 = ?

T1 = 35°C = 35 + 273 = 308K

T2 = 273K (STP)

Using P1V1/T1= P2V2/T2

705 × 3/308 = 760 × V2/273

2115/308 = 760V2/273

Cross multiply

308 × 760V2 = 2115 × 273

234,080V2 = 577,395

V2 = 577,395 ÷ 234,080

V2 = 2.47L

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T_2=-125.58\°C

Explanation:

Hello!

In this case, considering the Gay-Lussac's law which describes the pressure-temperature behavior as a directly proportional relationship by holding the volume as constant, we write:

\frac{T_1}{P_1} =\frac{T_2}{P_2}

Whereas solving for the final temperature T2, we get:

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