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sertanlavr [38]

3 years ago

5

What is the approximate energy required to raise the temperature of 1.00 L of hydrogen by 90 °C? The pressure is held constant a

nd equal to 1 atm.

1 answer:

Oliga [24]3 years ago

3 0

Answer:

Q = 116.8 J

Explanation:

Here given that the temperature of 1 L hydrogen is increased by 90 degree C at constant pressure condition.

So here we will have

$Q = n C_p \Delta T$

here we know that

n = number of moles

$n = \frac{1}{22.4}$

$n = 0.0446$

for ideal diatomic gas molar specific heat capacity at constant pressure is given as

$C_p = \frac{7}{2}R$

now we have

$Q = (0.0446)(\frac{7}{2}R)(90)$

$Q = 116.8 J$

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A rigid adiabatic container is divided into two parts containing n1 and n2 mole of ideal gases respectively, by a movable and th

kicyunya [14]

Answer:

Explanation:

Given

Pressure, Temperature, Volume of gases is

$P_1, V_1, T_1 & P_2, V_2, T_2$

Let P & T be the final Pressure and Temperature

as it is rigid adiabatic container therefore Q=0 as heat loss by one gas is equal to heat gain by another gas

$-Q=W+U_1----1$

$Q=-W+U_2-----2$

where Q=heat loss or gain (- heat loss,+heat gain)

W=work done by gas

$U_1 & U_2$ change in internal Energy of gas

Thus from 1 & 2 we can say that

$U_1+U_2=0$

$n_1c_v(T-T_1)+n_2c_v(T-T_2)=0$

$T(n_1+n_2)=n_1T_1+n_2T_2$

$T=\frac{n_1+T_1+n_2T_2}{n_1+n_2}$

where $n_1=\frac{P_1V_1}{RT_1}$

$n_2=\frac{P_2V_2}{RT_2}$

$T=\frac{\frac{P_1V_1}{RT_1}\times T_1+\frac{P_2V_2}{RT_2}\times T_2}{\frac{P_1V_1}{RT_1}+\frac{P_2V_2}{RT_2}}$

$T=\frac{P_1V_1+P_2V_2}{\frac{P_1V_1}{T_1}+\frac{P_2V_2}{T_2}}$

and $P=\frac{P_1V_1+P_2V_2}{V_1+V_2}$

6 0

3 years ago

John decided to cycle to his friend's house at a speed of 5km/h and the journey took 2

Marianna [84]

Answer:100 miles

Explanation:

5 0

3 years ago

Read 2 more answers

A rocket moves upward, starting from rest with an acceleration of 25.4 m/s^2 for 3.39 s. It runs out of fuel at the end of the 3

kolbaska11 [484]

Explanation:

Initial speed of the rocket, u = 0

Acceleration of the rocket, $a=25.4\ m/s^2$

Time taken, t = 3.39 s

Let v is the final velocity of the rocket when it runs out of fuels. Using the equation of kinematics as :

$v=u+at$

$v=25.4\times 3.39=86.10\ m/s$

Let x is the initial position of the rocket. Using third equation of kinematics as :

$v^2=u^2+2ax_o$

$x_o=\dfrac{v^2}{2a}$

$x_o=\dfrac{86.10^2}{2\times 25.4}=145.92\ m$

Let $x_o$ is the position at the maximum height. Again using equation of motion as :

$v^2-u^2=2a(x-x_o)$

Now $a=-g$ and v and u will interchange

$u^2=2g(x-x_o)$

$x=x_o+\dfrac{u^2}{2g}$

$x=145.92+\dfrac{(86.10)^2}{2\times 9.8}$

x = 524.14 meters

Hence, this is the required solution.

5 0

3 years ago

A red laser with a wavelength of 670 nm and a blue laser with a wavelength of 450 nm emit laser beams with the same light power.

tatyana61 [14]

E=hf C=wavelength*F

E=hC/wavelength

E=(6.626*10^-34)*(3.00*10^8)/670*10^-9

E=(6.626*10^-34)*(3.00*10^8)/450*10^-9

6 0

3 years ago

The speed of sound is measured to be 340 m/s on a certain day.

ziro4ka [17]

Answer: 1,224 km/h

Explanation:

To do this, we pick the first unit and convert
Picking m first and converting to km:
Since we're converting from a non-prefix to a prefix, we divide the value by the prefix were taking it to. In this case, kilo = 10³ which means we're going to divide our value by 1000 to convert it from m to km
340 m/s ÷ 1000 = 0.34 km/s
Now, let's convert our seconds to hour:
We'll need to calculate how many hours is equivalent to one second first;
1 hr = 60×60 seconds
X hr = 1 second
*Cross multiply*
1 × 1 = X × 60 × 60
1 = 3,600 X
X = 1 / 3,600
X = 2.778×10⁻⁴ hour
So, in the place of "1 Second", we're going to be inserting 2.778×10⁻⁴ hour instead
0.34 km / s = 0.34 km / 2.778×10⁻⁴ hour
(0.34 / 2.778×10⁻⁴) km/hour
1,224 km/h.
340 m/s = 1,224 km/h

6 0

2 years ago