The pressure and volume of an expanding gas are related by the formula PV^b=C, where b and C are constants (this holds in adiaba
1 answer:
The dP/dt of the adiabatic expansion is -42/11 kPa/min
<h3>How to calculate dP/dt in an adiabatic expansion?</h3>An adiabatic process is a process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression
Given b=1.5, P=7 kPa, V=110 cm³, and dV/dt=40 cm³/min
PVᵇ = C
Taking logs of both sides gives:
ln P + b ln V = ln C
Taking partial derivatives gives:
Substitutituting the values b, P, V and dV/dt into the derivative above:
1/7 x dP/dt + 1.5/110 x 40 = 0
1/7 x dP/dt + 6/11 = 0
1/7 x dP/dt = - 6/11
dP/dt = - 6/11 x 7
dP/dt = -42/11 kPa/min
Therefore, the value of dP/dt is -42/11 kPa/min
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Answer:
Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box has largest volume.
Step-by-step explanation:
Given that,
A box is being created out of a 15 inches by 10 inches sheet of metal.
The length of the one side of the squares which are cut out of the each corners of the metal sheet be x.
The length of the metal box be = (15-2x) inches.
The width of the metal box be =(10-2x) inches
The height of the metal box be =x inches
Then, the volume of the metal box= length×width×height
=(15-2x)(10-2x)x cubic inches
=(150x-50x²+4x³) cubic inches
∴ V= 4x³-50x²+15x
Differentiating with respect to x
V'=12x²-100x+15
Again differentiating with respect to x
V''=24x-100
For maximum or minimum value, V'=0
12x²-100x+15=0
Apply quadratic formula , here a=12, b= -100 and c=15
For x= 8.18, The value of (15-2x) and (10-2x) will negative.
∴x=0.1528 .
Now,
∴At x=0.1528 inch, the volume of the metal box will be maximum.
Therefore the dimensions of the square should be 0.1528 inch by 0.1528 inch so, the box has largest volume.