a balloon containing 0.500 mol ar at 0.00°c and 65.0 kpa pressure is expanded by adding more argon. formula…

a balloon containing 0.500 mol ar at 0.00°c and 65.0 kpa pressure is expanded by adding more argon. formula: $pv = nrt$ ($r = 8.31 lcdot kpa/molcdot k$) what is the original volume of the gas?

a balloon containing 0.500 mol ar at 0.00°c and 65.0 kpa pressure is expanded by adding more argon. formula: $pv = nrt$ ($r = 8.31 lcdot kpa/molcdot k$) what is the original volume of the gas?

Answer

Explanation:

Step1: Convert temperature to Kelvin

$T = 0.00^{\circ}C+273.15 = 273.15\ K$

Step2: Rearrange the ideal - gas law for volume

$V=\frac{nRT}{P}$

Step3: Substitute the given values

$n = 0.500\ mol$, $R = 8.31\ L\cdot kPa/mol\cdot K$, $T = 273.15\ K$, $P = 65.0\ kPa$ $V=\frac{0.500\ mol\times8.31\ L\cdot kPa/mol\cdot K\times273.15\ K}{65.0\ kPa}$ $V=\frac{0.500\times8.31\times273.15}{65.0}\ L$ $V = 17.7\ L$

Answer:

$17.7$