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Answer :
Final Answer:
The temperature of the gas at standard pressure is C) 66.23 °C,(Option.c)
Explanation:
Gay-Lussac's law states that at constant volume, the pressure of a gas is directly proportional to its temperature. Mathematically, this relationship is expressed as P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 and T2 are the final pressure and temperature, respectively.
Given:
Initial pressure (P1) = 699.0 torr
Initial temperature (T1) = 40 °C
We need to find the temperature at standard pressure, which is 1 atm (760 torr). We can denote this as P2.
Converting the temperatures to Kelvin scale:
T1 = 40 °C + 273.15 = 313.15 K
Plugging the values into the Gay-Lussac's law equation:
699.0 torr / 313.15 K = 760 torr / T2
Cross multiplying and solving for T2:
699.0 torr * T2 = 760 torr * 313.15 K
T2 = (760 torr * 313.15 K) / 699.0 torr
T2 ≈ 339.38 K
Converting the temperature back to Celsius:
T2 ≈ 339.38 K - 273.15 ≈ 66.23 °C
Therefore, the correct temperature of the gas at standard pressure is approximately C) 66.23 °C,(Option.c)
Question
At a pressure of 699.0 torr, a gas has a temperature of 40 °C. What is the temperature of the gas at standard pressure (1 atm = 760 torr)?
A) 40 °C
B) 273 °C
C) 66.23 °C,
D) 313 °C
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