Question

Initially, a 55.0 liter compressible container, holding 2.4 moles of a gas, exerts a pressure of 760 millimeters of mercury at a temperature of 280 Kelvin. What is the pressure when the container is compressed to 43.0 liters, the moles of gas reduces to 1.8 moles, and the temperature changes to 36 degrees Celsius?
93.7 mm Hg
492 mm Hg
740 mm Hg
805 mm Hg

Answer 1

You would have to use the ideal gas law for this:
PV=nRT
Pressure, Volume, n=moles, R gas constant, Temperature in Kelvin
P=nRT/V
(1.8mol)(62.36)(309K)/43.0L = 805mm Hg 

Answer 2

Answer:

805 mmHg is the pressure when the container is compressed

Explanation:

Initial Volume of the gas, V = 55.0 L

Initial Moles of the gas, n = 2.4 mol

Initial Pressure of the gas ,P=760 mmHg

initial temperature of the gas, T = 280 K

Using an Ideal gas equation:

$PV=nRT$

$R=\frac{PV}{nT}$...(1)

Final Volume of the gas, V '= 43.0 L

Final Moles of the gas, n' = 1.8 mol

Final Pressure of the gas = p'

Final temperature of the gas, T' =36 °C = 309 K

Using an Ideal gas equation:

$R=\frac{P'V'}{n'T'}$..(2)

(1)=(2)

$\frac{PV}{nT}=\frac{P'V'}{n'T'}$

$\frac{760 mmHg\times 55.0 L}{2.4 mol\times 280 K}=\frac{P'\times 43.0 L}{1.8 mol\times 309 K}$

P' = 805 mmHg

805 mmHg is the pressure when the container is compressed.