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If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ∘C , what is the difference between the ideal pressure (as predict

Rudik [331]

Answer:

1.98 atm

Explanation:

Given that:

Temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28 + 273.15) K = 301.15 K

n = 1

V = 0.500 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

P × 0.500 L = 1 ×0.0821 L atm/ K mol × 301.15 K

⇒P (ideal) = 49.45 atm

Using Van der Waal's equation

$\left(P+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$

R = 0.0821 L atm/ K mol

Where, a and b are constants.

For Ar, given that:

So, a = 1.345 atm L² / mol²

b = 0.03219 L / mol

So,

$\left(P+\frac{1.345\times \:1^2}{0.500^2}\right)\left(0.500-1\times 0.03219\right)=1\times 0.0821\times 301.15$

$P+\frac{1.345}{0.25}=\frac{24.724415}{0.46781}$

$P=\frac{24.724415}{0.46781}-\frac{1.345}{0.25}$

⇒P (real) = 47.47 atm

Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm

4 0

2 years ago

Two cables of the same length are made of the same material, except that one cable has twice the diameter of the other cable. Wh

belka [17]

Answer: The same current flows through bth cables

Explanation:

Lets have a look to the next two equations

The Ohm´s V = I*R (1)

where:

V is voltage (potencial dfference) in volts

I is the electric current in ampers

R is the electric resistance

When a voltage is applied as the electrc load is not specified ( we have to assume is the same) the current will be the same

And in the other hand the resistance R =ρL/s

Where ρ is the resistivity of the conductor L the length and s square section of the conductor

If we assume that the smaller diameter cable is able to conduct the current then nothing happens. The point is that the capacity of conduction of current depend on the section of the cable (the area)

Tables exist where to find the capacity of each cable according to its diameter.

6 0

2 years ago

A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7

maxonik [38]

First, let's find the speed $v_i$ of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
$p_i = m_1 v_1$
After the collision, the two blocks stick together and so now they have mass $m_1 +m_2$ and they are moving with speed $v_i$ :
$p_f = (m_1 + m_2)v_i$
For conservation of momentum
$p_i=p_f$
So we can write
$m_1 v_1 = (m_1 +m_2)v_i$
From which we find
$v_i = \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s$

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force $\mu (m_1+m_2)g$ . The work done by the frictional force to stop the two blocks is
$\mu (m_1+m_2)g d$
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
$\frac{1}{2} (m_1+m_2)v_i^2$
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
$\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g d$
From which we can find the value of the coefficient of kinetic friction:
$\mu = \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49$

3 0

2 years ago

Th e heat capacity of air is much smaller than that of water, and relatively modest amounts of heat are needed to change its tem

Darina [25.2K]

Answer:

Q=1005 J

t= 0.67 sec

Explanation:

Lets take condition of room is 1 atm and 25°C.

Heat capacity ,c = 21 J /K.mol

If we assume that air is ideal gas that

P V = n R T

$V= 5.5\times 6.5\times 3\ m^3$

$V=107.25\ m^3$

$V=107.25\times 1000 L$

V= 107250 L

At STP number of moles given as

$n=\dfrac{V}{V_{at\ S.T.P.}}$

V=22.4 L at S.T.P.

$n=\dfrac{107250}{22.4}\ moles$

n=4787.94 moles

n= 4.784 Kmoles

So heat required to raise 10°C temperature

Q = n x c x ΔT

Q = 4.78794 x 21 x 10

Q=1004.64 J

Time t

t= Q/P

P= 1.5 KW

t = 1.004.64 /1.5

t= 0.66 sec

4 0

2 years ago

Read 2 more answers

Which trailer has more downward pressure where it attaches to the car?

VARVARA [1.3K]

The one that is loaded worst. The overall weight is not important; tongue weight is a matter of loading. Our 12,000 lb snow cat trailer, which has stops to position the cat properly, has under 100 lbs tongue weight. Excessive tongue weight is a Bad Thing because it reduces weight on the towing vehicle's front wheels, leading to instability.

4 0

2 years ago

Any person who opens the door he applies​

Any person who opens the door he applies