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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

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dudepurple0

Answer:

Answer is c....115°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

Lanuel

Based on the calculations, the new temperature of this ideal gas at constant volume is equal to: C. 115°C.

Given the following data:

  • Initial temperature = 25°C to K = 273 + 25 = 298 K.
  • Final pressure = 1.25 atm.
  • Initial pressure = 0.96 atm.

How to determine the new temperature?

In order to calculate the new temperature of this ideal gas at constant volume, we would apply Gay Lussac's law.

Mathematically, Gay Lussac's law is given by this formula;

P α T

P = kT

Where:

  • P is the pressure.
  • T is the temperature.
  • k is the constant of proportionality.

Substituting the given parameters into the formula, we have;

[tex]\frac{P_1}{T_1} =\frac{P_2}{T_2} \\\\\frac{0.96}{298} =\frac{1.25}{T_2} \\\\0.96T_2 =372.5[/tex]

T₂ = 372.5/0.96

New temperature = 388.0 K.

Conversion:

°C = K - 273

°C = 388.0 - 273

New temperature = 115°C.

Read more on temperature here: https://brainly.com/question/888898

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