Question

A cylinder with a moveable piston holds 3.60 mol of argon at a constant temperature of 295 K. As the gas is compressed isothermally, its pressure increases from 101 kPa to 151 kPa. (a) Find the final volume of the gas.

Answer #1

Universal gas constant = R = 8.314 J/(mol.K)

Number of moles of argon gas = n = 3.6 mol

Initial temperature of the argon gas = T_{1} = 295 K

Initial pressure of the argon gas = P_{1} = 101 kPa =
101 x 10^{3} Pa

Initial volume of the argon gas = V_{1}

By Ideal Gas Law,

P_{1}V_{1} = nRT_{1}

(101x10^{3})V_{1} = (3.6)(8.314)(295)

V_{1} = 8.742 x 10^{-2} m^{3}

Final pressure of the argon gas = P_{2} = 151 kPa = 151
x 10^{3} Pa

Final volume of the argon gas = V_{2}

The process is isothermal therefore the temperature remains constant.

P_{1}V_{1} = P_{2}V_{2}

(101x10^{3})(8.742x10^{-2}) =
(151x10^{3})V_{2}

V_{2} = 5.85 x 10^{-2} m^{3}

a) Final volume of the argon gas = 5.85 x 10^{-2}
m^{3}

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