The given question is incomplete. The complete question is :
You fill a tank with gas at 60C and 100 kPa and seal it. You decrease the temperature to 10C but keep the volume constant. What happens to the pressure?
Answer: Thus the pressure will decrease to 85.0 kPa
Explanation:
Gay-Lussac's Law : It is defined as the pressure is directly proportional to the temperature of the gas at constant volume and number of moles.
(Volume, moles of gas are constant)
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=100kPa\\T_1=60^oC=(60+273)K=333K\\P_2=?\\T_2=10^0C=(10+273)K=283K[/tex]
Putting values in above equation, we get:
[tex]\frac{100}{333}=\frac{P_2}{283}\\\\P_2=85.0kPa[/tex]
Thus the pressure will decrease to 85.0 kPa
