Question 3 (11 points) A gas has a volume of 690.0mL at -15.1°C and 392.0 mmHg. What would the volume of the gas be at
233.0°C and 0.700 atm of pressure? Answer with no decimal places.

1 L

Explanation:

The ideal gas law is PV = nRT. However, pressure has to be atm, volume has to be in L, and temperature has to be in K. So let's convert. (R is the gas constant 0.0821)

P = 392.00 mmHg

760 mmHg = 1 atm

392.00 mmHg x = 0.516 atm

V = 690.0 mL

1 L = 1000 mL

690.0 mL × = 0.69 L

T = -15.1° C

K = °C + 273

K = -15.1° C + 273 = 258 K

Now, plug in the numbers into the ideal gas law equation to find the number of moles.

PV = nRT

(0.516)(0.69) = n(0.0821)(258)

0.35604 = 21.1818n

n = 0.0168 moles

Now, the formula for the volume of a gas is V = nRT/P. Plug in the information to solve for the volume of the gas.

T = 233.0° C

K = 233.0° C + 273 = 506 K

V = nRT ÷ P

V = (0.0168)(0.0821)(506) ÷ 0.700

V = 0.69791568 ÷ 0.700

V = 0.9970224 = 1 L

The volume of the gas is 1 L or 1000 mL.

Hope this helps. :)

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they're all from the same group as sulfur (group 16).