A gas in a rigid container at 25°C has a pressure of 0.96 atm. A change in temperature causes the pressure to increase to 1.25 atm. What is the new temperature of the gas? Use StartFraction P subscript 1 over T subscript 1 EndFraction equals StartFraction P subscript 2 over T subscript 2 EndFraction.. –44.2°C 32.6°C 115°C 388°C

Answer is c115°C

Explanation:

Gay-Lussacs law says that pressure of a gas is the same as temperature when the volume is kept the same

P / T = k

where P - pressure , T - temperature in kelvin and k - constant

parameters for the first time are on the left side of the equation and parameters for the second instance are on the right side of the equation

T1 - 25 °C + 273 = 298 K

substiuting the values in the equation

T2 = 388 K

temperature in celcius - 388 K - 273 = 115 °C