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

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.