Question

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

Answer 1

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

Answer 2

According to the combined gas laws,

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

`P_(1) = 0.983 atm`

`V_(1)=285 mL`

`T_(1) = 15^(0) + 273 = 288 K`

`P_(2) = ?`

`V_(2)=292 mL`

`T_(2) = 11^(0) + 273 = 284 K`

Plugging in the values and solving for the final pressure:

`(0.983 atm(285 mL))/(288 K) = (P_(2)(292 mL))/(284 K)`

Final pressure
`P_(2) = 0.946 atm`