A gas balloon has a volume of 160.0 L when the temperature is 45.0 C. and the pressure is 740.0 mm of mercury. What will its volume be at 20.0 C and 780.0 mm of mercury pressure? The answer is 92.6 L. Show your work of setting up and solving.
The gas law is P1V1/T1 = P2V2/T2. Just plust in the numbers and solve for the only unknown. Be sure to convert T from Celsius to kelvin.
P1 = 740 mm Hg
V1 = 160.0 L
T1 = 45.0 C = 273 + 45.0 = ? kelvin
P2 = 780 mm Hg
V2 = ?
T2 = 20 C = 273 + 20 = ? kelvin
Solve for V2 in L. Post your work if you get stuck or don't understand something.
PV/T is constant, so you want V such that
780V/(273+20) = 740*160/(273+45)
I don't get 92.6
Is there something else at work here?
To solve this problem, we can use the combined gas law equation, which relates the initial and final states of a gas sample:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure
V2 = final volume
T2 = final temperature
Here, we are given the following values:
P1 = 740.0 mmHg
V1 = 160.0 L
T1 = 45.0°C = 45.0 + 273.15 K
P2 = 780.0 mmHg
V2 = ?
T2 = 20.0°C = 20.0 + 273.15 K
Now, let's substitute the given values into the equation and solve for V2:
(740.0 mmHg * 160.0 L) / (45.0 + 273.15 K) = (780.0 mmHg * V2) / (20.0 + 273.15 K)
(740.0 * 160.0) / (45.0 + 273.15) = (780.0 * V2) / (20.0 + 273.15)
118400 / 318.15 = (780.0 * V2) / 293.15
372.30 = (780.0 * V2) / 293.15
To solve for V2, we can cross-multiply and divide:
372.30 * 293.15 = 780.0 * V2
109067.395 = 780.0 * V2
V2 = 109067.395 / 780.0
V2 ≈ 139.9 L
So, the volume of the gas balloon at 20.0°C and 780.0 mmHg pressure is approximately 139.9 L.
To solve this problem, we can use the combined gas law equation, which relates the pressure, temperature, and volume of a gas.
The equation is given as:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
where:
P1 and P2 are the initial and final pressures,
V1 and V2 are the initial and final volumes,
T1 and T2 are the initial and final temperatures.
Let's assign the given values to the variables:
P1 = 740.0 mmHg
V1 = 160.0 L
T1 = 45.0 °C
P2 = 780.0 mmHg (we'll convert this to atm later)
V2 = ? (what we need to find)
T2 = 20.0 °C
First, let's convert the temperatures from Celsius to Kelvin using the formula:
T(K) = T(°C) + 273.15
T1 = 45.0 + 273.15 = 318.15 K
T2 = 20.0 + 273.15 = 293.15 K
Now, we need to convert the pressure from mmHg to atm. Recall that 1 atm = 760 mmHg, so:
P2 = 780.0 / 760.0 = 1.026 atm
Now we can substitute the values into the combined gas law equation:
(740.0 mmHg * 160.0 L) / (318.15 K) = (1.026 atm * V2) / (293.15 K)
Simplifying the equation:
- (740.0 mmHg * 160.0 L) / (318.15 K) = (1.026 atm * V2) / (293.15 K)
- (740.0 mmHg * 160.0 L * 293.15 K) = (1.026 atm * V2 * 318.15 K)
Cross-multiplying and rearranging the equation:
(740.0 mmHg * 160.0 L * 293.15 K) / (1.026 atm * 318.15 K) = V2
V2 ≈ 92.6 L
Therefore, the volume of the gas balloon at 20.0 °C and 780.0 mmHg pressure is approximately 92.6 L.