Question

A sample of argon gas occupies a volume of 8.89 L at 49.0°C and 0.600 atm.

If it is desired to decrease the volume of the gas sample to 5.95 L, while increasing its pressure to 0.763 atm, the temperature of the gas sample at the new volume and pressure must be what in degrees celsius?

Answer 1

$\dfrac{P_{1}V_{1}} {T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}$

$\dfrac{8.89*0.6}{49+273=322}=\dfrac{5.95*0.763}{T_{2}}$

$5.334\ T_2=1461.8\newline T_2=\dfrac{1461.8}{5.334}$

$T_2= 274k$

convert $T_2$ into $Celsius$

$274k-273=\ 1$

Therefore,

$T_2=1^0\ C$