A hydrogen gas in piston/cylinder assembly at 280K, 100kPa with a volume of 0.1 m^3 is now compressed to a volume of 0.01 m^3 in a reversible adiabatic process. Find:
a) the new temperature, in K
b) how much work is required, kJ/kg
a) the new temperature, in K
b) how much work is required, kJ/kg
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This is hydrogen gas, which is diatomic (H2). This means that the adiabatic index, gamma, of the substance is 7/5. We know for adiabatic processes that
PV^gamma = constant (c). We know the initial pressure and volume, so let's find the constant.
100*0.1^(7/5) = 3.981 = c
We have a new volume, so let's find the new pressure.
P2*0.01^(7/5) = 3.981. P2 = 2512 kPa.
To find the new temp (a), we can use the Combined Gas Law. P1V1/T1 = P2V2/T2.
(100*0.1)/280 = (2512*0.01)/T2; T2 = 703K.
To answer part b, we need to use the knowledge that W = INTEGRAL(PdV), or the area under the PV curve. Since P and T change, I find it better to use a simpler integral:
W = c*INTEGRAL(dV/V^gamma). Let's evaluate the integral:
W = 3.981*INTEGRAL(V^(-7/5))dV, evaluated from 0.01 to 0.1.
W = 3.981*(V^(-2/5)/(-2/5)), evaluated from 0.01 to 0.1.
W = 3.981*9.49 = 37.80 kJ.
However, you're looking for units of kJ/kg. We have to know how much mass of hydrogen gas there is. Let's use the Ideal Gas Law to figure out how many moles of gas:
PV = nRT
(100)(0.1) = n(8.314)(280)
n = 0.004296 moles = 0.00866 grams = 8.66 x 10^-6 kg.
So the work density must be (37.8/(8.66 x 10-6)) = 4.36 x 10^6 kJ/kg.
I'm doing this without a pencil and paper so I hope I haven't made any stupid math errors. The way to go about this is right though.
PV^gamma = constant (c). We know the initial pressure and volume, so let's find the constant.
100*0.1^(7/5) = 3.981 = c
We have a new volume, so let's find the new pressure.
P2*0.01^(7/5) = 3.981. P2 = 2512 kPa.
To find the new temp (a), we can use the Combined Gas Law. P1V1/T1 = P2V2/T2.
(100*0.1)/280 = (2512*0.01)/T2; T2 = 703K.
To answer part b, we need to use the knowledge that W = INTEGRAL(PdV), or the area under the PV curve. Since P and T change, I find it better to use a simpler integral:
W = c*INTEGRAL(dV/V^gamma). Let's evaluate the integral:
W = 3.981*INTEGRAL(V^(-7/5))dV, evaluated from 0.01 to 0.1.
W = 3.981*(V^(-2/5)/(-2/5)), evaluated from 0.01 to 0.1.
W = 3.981*9.49 = 37.80 kJ.
However, you're looking for units of kJ/kg. We have to know how much mass of hydrogen gas there is. Let's use the Ideal Gas Law to figure out how many moles of gas:
PV = nRT
(100)(0.1) = n(8.314)(280)
n = 0.004296 moles = 0.00866 grams = 8.66 x 10^-6 kg.
So the work density must be (37.8/(8.66 x 10-6)) = 4.36 x 10^6 kJ/kg.
I'm doing this without a pencil and paper so I hope I haven't made any stupid math errors. The way to go about this is right though.