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user100 [1]
2 years ago
13

What is the effect of the following change on the volume of 1 mol of an ideal gas? The initial pressure is 722 torr, the final p

ressure is 0.950 atm, the initial temperature is 32°F, and the final temperature is 273 K. A. Volume is unchanged B. Volume increases by about 10% C. volume decreases by about 92% D. volume increases by 8% E. volume increases by about 90%.
Physics
1 answer:
Ray Of Light [21]2 years ago
4 0

Answer:

A. Volume is unchanged

Explanation:

P_{i} = initial pressure of the gas = 722 torr = 96258.7 pa

P_{f} = final pressure of the gas = 0.950 atm = 96258.75 pa

T_{i} = initial temperature = 32 °F = 272.15 K

T_{f} = final temperature = 273 K

V_{i} = initial volume

V_{f} = final volume

Using the Equation

\frac{P_{i} V_{i}}{T_{i}} = \frac{P_{f} V_{f}}{T_{f}}

Inserting the values

\frac{(96958.7) V_{i}}{272.15} = \frac{(96958.75) V_{f}}{273}

V_{f} = (1.00312) V_{i}

Hence the volume is unchanged.

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Chris and Jamie are carrying Wayne on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 100 N. Wayne weigh
PIT_PIT [208]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The value is F_j  =  550\ N

Explanation:

From the question we are told that

   The length of the stretcher is  d =  2.0 \  m

    The weight of the stretcher is W  =  100 \  N

    The weight for Wayne is  W_w =  800 \ N

     The distance of  center of gravity for Wayne from Chris is c_w = 75 cm  =  0.75 \ m

Generally taking moment about the first end where Chris is

         F_j *  d              => upward moment

Here F_j is the force applied by Jamie

Generally  taking moment about the second end where Jamie is

      W *  ( \frac{d}{2} ) +  W_w * (d - c_w)      => downward moment

Generally at equilibrium , the upward moment is equal to the downward moment

     F_j *  d = W *  ( \frac{d}{2} ) +  W_w * (d - c_w)

=>   F_j *  2  = 100 *  ( \frac{ 2}{2} ) +  800 * (2 - 0.75)

=>    F_j  =  550\ N

3 0
2 years ago
Gretchen runs the first 4.0 km of a race at 5.0 m/s. Then a stiff wind comes up, so she runs the last 1.0 km at only 4.0 m/s.
KIM [24]

Answer:

The velocity is v = 4.76 \ m/s

Explanation:

From the question we are told that

   The first distance is   d_1  =  4.0 \ km  =  4000 \ m

   The  first speed  is  v_1 =  5.0 \ m/s

    The  second distance is  d_2  =  1.0 \ km  =  1000 \ m

    The  second speed  is  v_2  =  4.0 \ m/s

Generally the time taken for first distance is  

      t_1 =  \frac{d_1 }{v_1 }

        t_1 =  \frac{4000}{5}

       t_1 =  800 \ s

The time taken for second  distance is

           t_1 =  \frac{d_2 }{v_2 }

        t_1 =  \frac{1000}{4}

       t_1 =  250 \ s

The total time is mathematically represented as

     t =  t_1 + t_2

=>   t =  800 + 250

=>    t =  1050 \ s

Generally the constant velocity that would let her finish at the same time is mathematically represented as

      v =  \frac{d_1 + d_2}{t }

=>    v =  \frac{4000 + 1000}{1050 }

=>    v = 4.76 \ m/s

7 0
2 years ago
Two workhorses tow a barge along a straight canal. Each horse exerts a constant force of magnitude F, and the tow ropes make an
morpeh [17]

Answer:

<em>a) Fvt cosθ</em>

<em>b) Fv cosθ</em>

<em></em>

Explanation:

Each horse exerts a force = F

the rope is inclined at an angle = θ

speed of each horse = v

a) In time t, the distance traveled d = speed x time

i.e d = v x t = vt

also, the resultant force = F cosθ

Work done W = force x distance

W = F cosθ x vt = <em>Fvt cosθ</em>

<em></em>

b) Power provided by the horse P = force x speed

P = F cosθ x v

P = <em>Fv cosθ</em>

8 0
2 years ago
A container of volume 0.6 m^3 contains 5.3 mol of argon gas at 24°C. Assuming argon behaves as an ideal gas, find the total inte
Vitek1552 [10]

Answer:

the internal energy of the gas is 433089.52 J

Explanation:

let n be the number of moles, R be the gas constant and T be the temperature in Kelvins.

the internal energy of an ideal gas is given by:

Ein = 3/2×n×R×T

     = 3/2×(5.3)×(8.31451)×(24 + 273)

     = 433089.52 J

Therefore, the internal energy of this gas is 433089.52 J.

5 0
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En la etiqueta de un bote de fabada de 350 g, leemos que su aporte energético es de 1630 kj por cada 100 g de producto a) La can
fredd [130]

Answer:

(a) 153.37 g

(b) 5705 kJ

Explanation:

(a) To find the amount of bean needed by a man you first calculate the equivalence in beans to 2500kJ

2500kJ*\frac{100g}{1630kJ}=153.37\ g

Thus, 153.37 g has the energy needed by a man that needs 200kJ per day.

(b) The amount of energy per pot of bean is given by:

E=350g*\frac{1630kJ}{100g}\\\\E=5705\ kJ

Thus, the energy is 5705kJ

6 0
2 years ago
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