Question

A sample of neon gas occupies a volume of 10.2 L at 1.8 atm and 22.3°C. What would its temperature (in Kelvin) be at 1.5 atm?

Answer 1

P1=1.8 P2=1.5 T1=295k. T2=?
P1/T1=P2=T2
18/2950=15/10T
T=15×2950/12
T=246(approximately)

Answer 2

Answer : The temperature at 1.5 atm is 246.083 K.

Solution : Given,

$P_1=1.8atm$ at $T_1=22.3^oC=273+22.3=295.3K$       $(1^oC=273K)$

$P_2=1.5atm$

According to the Gay-Lussac's law, the pressure of gas is directly proportional to the temperature of the gas at constant volume.

$P\propto T\\\frac{P}{T}=constant\\\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where, P and T are the pressure and temperature respectively.

By rearranging the above formula, we get

${T_2}=\frac{P_2\times T_1}{P_1}$

Now put all the given values in this formula.

${T_2}=\frac{1.5atm\times 295.3K}{1.8atm}=246.083K$

Therefore, the temperature at 1.5 atm is 246.083 K.