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? Use StartFraction P subscript 1 over T subscript 1 EndFraction equals StartFraction P subscript 2 over T subscript 2 EndFraction. â€""44. 2°C 32. 6°C 115°C 388°C.

Answer 1

The new temperature of the gas is 115 °C

From the question given above, the following data were obtained:

• Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
• Initial pressure (P₁) = 0.96 atm
• New pressure (P₂) = 1.25 atm
• New temperature (T₂) =?

The new temperature of gas can be obtained as follow:

$\frac{P_1}{T_1} = \frac{P_2}{T_2} \\ \\ \frac{0.96}{298} = \frac{1.25}{T_2} \\ \\ cross \: multiply \\ \\ T_2 \times 0.96 = 298 \times 1.25 \\ \\ divide \: both \: side \: by \: 0.96 \\ \\T_2 = \frac{298 \times 1.25 }{0.96} \\ \\ T_2 = 388 \: K \\ \\ subtract \: 273 \: from \: 388 \: to \: express \: the \: answer \: im \: \degree \: C \\ \\ T_2 = 388 - 273 \\ \\ T_2 = 115 \: \degree \: C$

Therefore, the new temperature is 115 °C

Learn more about gas laws: brainly.com/question/25862018