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Andrej [43]
3 years ago
15

The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pre

ssure is maintained constant at 160 kPa. Determine: (a) The initial and final temperatures of helium. (b) The work required to compress it, in kJ.
Physics
1 answer:
IrinaVladis [17]3 years ago
6 0

Answer:

A) T1 = 269.63 K

T2 = 192.59 K

B) W = -320 KJ

Explanation:

We are given;

Initial volume: V1 = 7 m³

Final Volume; V2 = 5 m³

Constant Pressure; P = 160 KPa

Mass; m = 2 kg

To find the initial and final temperatures, we will use the ideal gas formula;

T = PV/mR

Where R is gas constant of helium = R = 2.0769 kPa.m/kg

Thus;

Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K

Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K

B) world one is given by the formula;

W = P(V2 - V1)

W = 160(5 - 7)

W = -320 KJ

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Which one of the following is the correct option for fill in the
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Answer:

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Enzyme reaction and deactivation Lipase is being investigated as an additive to laundry detergent for removal of stains from fab
natita [175]

Answer:

t= 27.38 mins [this the time taken by the enzyme to hydrolyse 80% of the fat present]

Explanation:given values

Half life of lipase t_1/2 = 8 min x 60s/min = 480 s

Rate constant for first order reaction

k_d = 0.6932/480 = 1.44 x 10^-3 s-1

Initial fat concentration S_0 = 45 mol/m3 = 45 mmol/L

rate of hydrolysis Vm0 = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration S = S_0(1-X) = 45 (1-0.80) = 9 mol/m3

K_m = 5mmol/L

time take is given by

t= -\frac{1}{K_d}ln[1-\frac{K_d}{V_m_0}(k_mln\frac{s_0}{s}+(s_0-s))]

all values are given and putting these value we get

t=1642.83 secs

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5 0
3 years ago
A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50
PSYCHO15rus [73]

Answer:

(a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

Explanation:

Given that,

Length = 50.00 cm

Time period = 2.50 s

Time period of pendulum is defined as the time for one complete cycle.

The period depends on the length of the pendulum.

Using formula of time period

T=2\pi\sqrt{\dfrac{I}{mgh}}

Where, I = moment of inertia

We need to calculate the value of d

Using parallel theorem of moment of inertia

I=I_{cm}+md^2

For a meter stick mass m , the rotational inertia about it's center of mass

I_{cm}-\dfrac{mL^2}{12}

Where, L = 1 m

Put the value into the formula of time period

T=2\pi\sqrt{\dfrac{\dfrac{mL^2}{12}+md^2}{mgd}}

T=2\pi\sqrt{\dfrac{L^2}{12gd}+\dfrac{d}{g}}

T^2=4\pi^2(\dfrac{L^2}{12gd}+\dfrac{d}{g})

Multiplying both sides by d

tex]T^2d=4\pi^2(\dfrac{L^2}{12g}+\dfrac{d^2}{g})[/tex]

(\dfrac{4\pi^2}{g})d^2-T^2d+\dfrac{\pi^2L^2}{3g}=0

Put the value of T, L and g into the formula

4.028d^2-6.25d+0.336=0

d = 0.056\ m, 1.496\ m

The value of d is 0.056 cm and 1.496 cm.

(b). Given that,

L = 50-5 = 45 cm

We need to calculate the time period

Using formula of period

T=2\pi\sqrt{\dfrac{l}{g}}

Put the value into the formula

T=2\pi\sqrt{\dfrac{45\times10^{-2}}{9.8}}

T=1.35\ sec

Hence, (a). The value of d is 0.056 cm and 1.496 cm.

(b). The time period is 1.35 sec.

7 0
3 years ago
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