Answer:
240 K, [tex]-33^{\circ}C[/tex]
Explanation:
For an ideal gas kept at constant volume, the pressure of the gas is proportional to its temperature. Mathematically,
[tex]\frac{p_1}{T_1}=\frac{p_2}{T_2}[/tex]
where in this problem
[tex]p_1 = 140 kPa[/tex] is the initial pressure of the gas
[tex]p_2 = 105 kPa[/tex] is the final pressure of the gas
[tex]T_1 = 47^{\circ} + 273 =320 K[/tex] is the initial absolute temperature of the gas
[tex]T_2[/tex] is the final absolute temperature of the gas
Solving for T2, we find:
[tex]T_2 = \frac{p_2 T_1}{p_1}=\frac{(105)(320)}{140}=240 K[/tex]
In degrees Celsius,
[tex]T_2 = 240 K - 273 = -33^{\circ}C[/tex]
