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

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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

\(\frac{P1}{T1} = \frac{P2}{T2}\)

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

\(\frac{0.96 atm}{298 K}= \frac{1.25 atm}{T2}\)

T2 = 388 K

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

answer is C. 115 °C