Final answer:
Using the combined gas law and the ideal gas constant, we can calculate the new temperature when a gas at 699.0 kPa and 40.0° C is changed to standard pressure of 101.3 kPa. First, the ideal gas constant, R, is calculated with known STP conditions, and then it's applied to find the temperature at the standard pressure.
Step-by-step explanation:
To calculate the temperature of gas at standard pressure, we use the ideal gas law which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature.
We are given that a gas has a pressure of 699.0 kPa at 40.0° C (which is 313.15 K when converted to Kelvin).
The student asks for the temperature at standard pressure (1 atm) which is equivalent to 101.3 kPa.
To find the new temperature at standard pressure, we will apply the combined gas law, assuming the volume and amount of gas (n) remain constant: (P1/T1 = P2/T2). First, we solve for the ideal gas constant R. Given 1 mol at STP, where P is 101.325 kPa, V is 22.414 L, and T is 273.15 K, we get:
R = PV / T
R = (101.325 kPa * 22.414 L) / 273.15 K
Then we use R to find the new temperature at standard pressure:
P1/T1 = P2/T2
699.0 kPa / 313.15 K = 101.3 kPa / T2
After cross-multiplying and solving for T2, we obtain the temperature at standard pressure.