A sample of argon has a pressure of 1.021 atm at a temperature of 89.4 degC. Assuming that the volume is constant, at what temperature will the gas have a pressure of 1016.56 mmHg
To solve this problem, we can use the ideal gas law equation, which states that the pressure (P) of a gas is directly proportional to its temperature (T), assuming volume (V) and amount of gas (n) are constant. The equation can be written 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.
To convert the given pressure from atm to mmHg, we can use the conversion factor:
1 atm = 760 mmHg
Now let's substitute the given values into the equation and solve for the final temperature (T2):
P1 = 1.021 atm
T1 = 89.4 °C
P2 = 1016.56 mmHg (converted to mmHg by multiplying by 760/1 atm/mmHg)
Let's plug the values into the equation and solve for T2:
P1/T1 = P2/T2
(1.021 atm) / (89.4 + 273.15) K = (1016.56 mmHg) / T2
To convert the initial temperature from Celsius to Kelvin, we add 273.15 to the Celsius value.
Now we need to cross-multiply and solve for T2:
T2 = (T1 * P2) / P1
T2 = (89.4 + 273.15) K * (1016.56 mmHg) / (1.021 atm)
T2 ≈ 375.55 K
Therefore, the gas will have a pressure of 1016.56 mmHg at a temperature of approximately 375.55 K.