Some hydrogen occupies 120 ft at 60.0°F. Find the temperature when its volume is 132ft^3 if the pressure remains constant.
At constant pressure,
V/T = constant. Use absolute Rankine temperature when using the equation. I will add 460 to the Fahrenheit temperatures to get degrees R
120/(460 + 60) = 132/(T + 460)
T + 460 = 132*520/120 = 572
T = 112 F
To find the temperature when the volume of hydrogen is 132 ft^3, we can use the ideal gas law equation:
PV = nRT
Where:
P = Pressure (constant)
V = Volume
n = Number of moles
R = Ideal gas constant
T = Temperature in Kelvin
First, we need to convert the given temperature from Fahrenheit to Kelvin using the formula:
T(K) = (T(°F) + 459.67) * (5/9)
Let's solve the given temperature at 60.0°F first:
T1(K) = (60.0 + 459.67) * (5/9)
= (519.67) * (5/9)
= 288.7 K
Now, we have:
V1 = 120 ft^3
T1 = 288.7 K
V2 = 132 ft^3
Since the pressure is constant, we can rewrite the equation as:
V1/T1 = V2/T2
Rearranging the equation to solve for the final temperature (T2):
T2 = (V2 * T1) / V1
Substituting the given values:
T2 = (132 * 288.7) / 120
≈ 318.2 K
Therefore, the temperature when the volume of hydrogen is 132 ft^3 (with constant pressure) is approximately 318.2 K.