Answer:
the pressure reading when connected a pressure gauge is 543.44 kPa
Explanation:
Given data
tank volume (V) = 400 L i.e 0.4 m³
temperature (T) = 25°C i.e. 25°C + 273 = 298 K
air mass (m) = 3 kg
atmospheric pressure = 98 kPa
To find out
pressure reading
Solution
we have find out pressure reading by gauge pressure
i.e. gauge pressure = absolute pressure - atmospheric pressure
first we find absolute pressure (p) by the ideal gas condition
i.e pV = mRT
p = mRT / V
p = ( 3 × 0.287 × 298 ) / 0.4
p = 641.44 kPa
so
gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 641.44 - 98
gauge pressure = 543.44 kPa
Answer:
1) the final temperature is T2 = 876.76°C
2) the final volume is V2 = 24.14 cm³
Explanation:
We can model the gas behaviour as an ideal gas, then
P*V=n*R*T
since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:
P*V^k = constant = C, k= adiabatic coefficient for air = 1.4
then the work will be
W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)
W = (P1*V1/T1)*(T2-T1)/(1-k)
T2 = (1-k)W* T1/(P1*V1) +T1
replacing values (W=-450 J since it is the work done by the gas to the piston)
T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C
the final volume is
TV^(k-1)= constant
therefore
T2/T1= (V2/V1)^(1-k)
V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³
