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?


a...–44.2°C

b....32.6°C

c....115°C

d...388°C

Respuesta :

W0lf93
Question: 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? a...–44.2°C b....32.6°C c....115°C d...388°C Answer: Answer is c....115°C.

Gay-Lussacs law states that pressure of a gas is directly proportional to temperature when the volume is kept constant

P / T = k

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

[tex] \frac{P1}{T1} = \frac{P2}{T2} [/tex]

where parameters for the first instance 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

substituting the values in the equation

[tex] \frac{0.96 atm}{298 K}= \frac{1.25 atm}{T2} [/tex]

T2 = 388 K

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

answer is C. 115 °C