a tank of ideal gas is sealed off at 20 degrees C and 1.00 atm pressure. What will be the pressure (in kPa and mmHg) in the tank if the gas temperature is decreased to -35degrees C ?
the book answer is
82kPa = 6.2 x10^2 mmHg
can you help me figure out how the book got this answer.
I've tried to plug the numbers into the gas formulas and have not had any luck.
At constant V and number of moles, P is proportional to absolute T.
P2/P1 = T2/T1 = 238/293 = 0.8123
P2 = 760 mm Hg * 0.8123 = 617 mm Hg
rRound it off to 620.
You have to know that 7600 mm Hg = 1 atmosphere, and how to convert degrees C to K
To solve this question, we can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's convert the temperatures from Celsius to Kelvin using the formula: K = °C + 273.15.
Given:
Initial temperature (T1) = 20 °C = 20 + 273.15 = 293.15 K
Initial pressure (P1) = 1.00 atm
Final temperature (T2) = -35 °C = -35 + 273.15 = 238.15 K
Final pressure (P2) = ?
Next, using the ideal gas law equation, we can rewrite it as:
P1V1 / T1 = P2V2 / T2
Assuming the volume (V) and the number of moles (n) remain constant, we can eliminate them from the equation, so the equation becomes:
P1 / T1 = P2 / T2
Now we can substitute the values:
P1 = 1.00 atm
T1 = 293.15 K
T2 = 238.15 K
Using the formula above, we can solve for P2:
P2 = (P1 * T2) / T1
P2 = (1.00 atm * 238.15 K) / 293.15 K
Calculating this, we find:
P2 ≈ 0.813 atm
Now, we can convert the pressure from atm to units of kPa and mmHg.
To convert atm to kPa, multiply the pressure by 101.325:
P2(kPa) = 0.813 atm * 101.325 kPa/atm
P2(kPa) ≈ 82.523 kPa ≈ 82 kPa (rounded to the nearest whole number)
To convert from atm to mmHg, multiply the pressure by 760:
P2(mmHg) = 0.813 atm * 760 mmHg/atm
P2(mmHg) ≈ 617.88 mmHg ≈ 620 mmHg (rounded to the nearest whole number)
Therefore, the book's answer of P2 = 82 kPa and 620 mmHg is correct.