A gas occupies 355 mL at a pressure of 99.5 kPa and a temperature of 22°C. The pressure
on the gas is subsequently increased by 10.0 kPa and the volume is dropped by 75 mL.
What's the new temperature?
-17*C
-17*C
To find the new temperature of the gas, we can use the combined gas law equation, which relates the initial and final conditions of pressure, volume, and temperature.
The combined gas law equation is:
(P1 * V1) / T1 = (P2 * V2) / T2
where:
P1 is the initial pressure,
V1 is the initial volume,
T1 is the initial temperature,
P2 is the final pressure,
V2 is the final volume,
T2 is the final temperature.
Given:
P1 = 99.5 kPa
V1 = 355 mL = 0.355 L
T1 = 22°C = 295 K (Since temperature must be in Kelvin)
Also, the pressure, P2 is increased by 10.0 kPa, and the volume, V2 is decreased by 75 mL.
To find P2 and V2, we need to add the change in pressure and subtract the change in volume from their respective initial values:
P2 = P1 + ΔP = 99.5 kPa + 10.0 kPa = 109.5 kPa
V2 = V1 - ΔV = 0.355 L - 0.075 L = 0.280 L (since volume must be in liters)
Now we can substitute these values into the combined gas law equation to solve for T2:
(P1 * V1) / T1 = (P2 * V2) / T2
(99.5 kPa * 0.355 L) / 295 K = (109.5 kPa * 0.280 L) / T2
103.94875 kPa * L / K = 30.66 kPa * L / T2
Cross multiplying and isolating T2:
103.94875 kPa * L / T2 = 30.66 kPa * L / 295 K
T2 = (30.66 kPa * L * 295 K) / (103.94875 kPa * L)
T2 ≈ 87.5 K
Therefore, the new temperature of the gas is approximately 87.5 K.