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2 years ago

4

find the kinetic energy of 5 litre of a gas at STP given standard pressure is 1.013 into 10 to the power 5 Newton per metre squa

re

1 answer:

Tems11 [23]2 years ago

7 0

Answer:

1.52 kJ

Explanation:

Total kinetic energy in gas K = 3NkT/2 where N = number of molecules, K = Boltzmann constant and T = temperature.

Also from the ideal gas law, PV = NkT

substituting PV into K above

K = 3PV/2 where P = pressure = 1.013 × 10⁵ N/m² and V = volume of gas = 5 litres = 5 dm³ = 5 × 10⁻³ m³

K = 3 × 1.013 × 10⁵ N/m² × 5 × 10⁻³ m³/2

= 1519.5 J

= 1.5195 kJ

≅ 1.52 kJ

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150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the wag

mafiozo [28]

Answer:

the friction force on this box is closest to 45.9 N.

Explanation:

given data

Weight of the box W = 150 N

accelerating uniformly = 3.00 m/s²

coefficient of kinetic friction = 0.400

coefficient of static friction = 0.600

solution

we know box does not move relative to the wagon

we get here friction force that is express as

friction force = mass × acceleration ..............1

here mass = weight ÷ g

mass = $\frac{150}{9.8}$ = 15.3 kg

put value in equation 1 we get

friction force = 15.3 kg × 3

friction force = 45.9 N

So, the friction force on this box is closest to 45.9 N.

3 0

2 years ago

A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of wa

gavmur [86]

Answer : The correct option is, (d) $535^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of copper = $0.10cal/g^oC$

$c_2$ = specific heat of water = $1.00cal/g^oC$

$m_1$ = mass of copper = 120 g

$m_2$ = mass of water = 300 g

$T_f$ = final temperature of mixture = $35^oC$

$T_1$ = initial temperature of copper = ?

$T_2$ = initial temperature of water = $15^oC$

Now put all the given values in the above formula, we get:

$120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC$

$T_1=535^oC$

Therefore, the temperature of the kiln was, $535^oC$

7 0

2 years ago

Read 2 more answers

An automobile is traveling at a constant 15 m/s, then it undergoes acceleration from that moment forward. Which statement best d

creativ13 [48]

Answer:

d

Explanation:

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2 years ago

A particle decelerates uniformly from a speed of 30 cm/s to rest in a time interval of 5.0 s. It then has a uniform acceleration

wel

Answer:

V=20cm/s

Explanation:

The average speed is the distance total divided the time total:

$V=X/T$

First stage:

T1=5s

$v_{f} =v_{o} - at$

But, $v_{f} =0$ (decelerates to rest)

then: $a =v_{o} /t=0.3/5=0.06m/s^{2}$

on the other hand:

$x =v_{o}*t - 1/2*at^{2}=0.3*5-1/2*0.06*5^{2}=0.75m$

X1=75cm

Second stage:

T2=5s

$x =v_{o}*t + 1/2*at^{2}=0+1/2*0.1*5^{2}=1.25m$

X2=125cm

Finally:

X=X1+X2=200cm

T=T1+T2=10s

V=X/T=20cm/s

8 0

2 years ago

A runner generates 1260 W of thermal energy. If this heat has to be removed only by evaporation, how much water does this runner

sergey [27]

Answer:0.502kg

Explanation:

F4om the relation

Power x time = mass x latent heat of vapourization

P.t=ML

1260 * 15 *60 = M * 22.6 * 10^5

M= 1134000/(22.6 *10^5)

M=0.502kg=502g

3 0

2 years ago