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A student collects 285 mL of O2 gas at a temperature of 15°C and a pressure of 0.983 atm. The next day, the same sample occupies 292 mL at a temperature of 11°C. What is the new pressure of the gas? 0.946 atm 1.00 atm 0.704 atm

Respuesta :

isyllus

Answer:-

0.946 atm

Explanation:-

From the question we see,

First day pressure P 1 = 0.983 atm

First day temperature T 1 =15 C + 273 = 288 K

First day volume V 1 =285 mL

Second Day volume V 2 = 292 mL

Second day Temperature T 2 = 11 C + 273 = 284 K

Using the relation

P 1 V 1 / T 1 = P 2 V 2 / T 2

We get Second Pressure P 2 = P 1 V 1 T 2 / ( T 1 V 2)

= 0.983 atm x 285 mL x 284 K / ( 288 K x 292 mL)

= 0.946 atm

Hence the new pressure of the gas on the second day is 0.946 atm

Chlidonias

According to the combined gas laws,

[tex] \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}} [/tex]

[tex] P_{1} = 0.983 atm [/tex]

[tex]V_{1}=285 mL [/tex]

[tex] T_{1} = 15^{0} + 273 = 288 K [/tex]

[tex] P_{2} = ? [/tex]

[tex] V_{2}=292 mL [/tex]

[tex] T_{2} = 11^{0} + 273 = 284 K [/tex]

Plugging in the values and solving for the final pressure:

[tex] \frac{0.983 atm(285 mL)}{288 K} = \frac{P_{2}(292 mL)}{284 K} [/tex]

Final pressure [tex] P_{2} = 0.946 atm [/tex]

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