Asked by angel
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
All Answers 1
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Use P1/T1 = P2/T2.
T1 = 25°C = 298.15 K.
T2 = T1*(P2/P1) = 298.15*(1.25/0.96) ≈ 388.25 K = 388.25 − 273.15 ≈ 115.1°C.
Answer: 115°C.
T1 = 25°C = 298.15 K.
T2 = T1*(P2/P1) = 298.15*(1.25/0.96) ≈ 388.25 K = 388.25 − 273.15 ≈ 115.1°C.
Answer: 115°C.
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