Question

A sample of methane occupies a volume of 174.0 mL at 25 C and exerts a pressure of 1030.0 mmHg if the volume of the gas is allowed to expand to 650.0 mL at 298K what will be the pressure of the gas?

Answer 1

Answer:

The pressure will be 0,37 atm.

Explanation:

We use the Boyle Mariotte´s formula,where for a given mass of gas, at constant temperature, the pressure and temperature vary inversely proportionally. We see here that the initial and final temperature does not vary (298K).

We convert the unit Celsius into Kelvin (0 ° C = 273K) and the unite mmHg into atm (760mmHg=1 atm):

25°C= 25+273=298K   ; (1060mmHgx 1 atm)/760mmHg)=1,39 atm.

We convert unit of volume from ml to L: 174/1000=0,174L and 650/1000=0,650L

P1xV1= P2xV2

1,39 atm x 0,174L =P2 x0,650L

P2= (1,39 atm x 0,174L)/0,650L= 0,37 atm

Answer 2

Answer:

The pressure of the gas will decrease to 275.7 mmHg or 0.362 atm

Explanation:

Step 1: Data given

Volume of methane = 174.0 mL = 0.174 L

Temperature = 25.0 °C = 298 K

Pressure = 1030 mmHg = 1030/760 atm = 1.355263 atm

Volume increases to 650.0 mL = 0.650 L

The temperature stays constant

Step 2: Calculate the new pressure

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 1.355263 atm

⇒with V1 = the initial volume = 0.174 L

⇒with P2 = the new pressure = TO BE DETERMINED

⇒with V2 = the new volume = 0.650 L

1.355263 * 0.174 = P2 * 0.650

P2 = 0.362 atm = 275.7 mmHg

OR

1030*0.174 = P2 * 0.650

P2 = 275.7 mmHg

The pressure of the gas will decrease to 275.7 mmHg or 0.362 atm