Physics, 07.12.2020 14:00 sydc1215

9. A constant mass of hydrogen occupies a volume of 4.0 cm at a pressure of 2.0 x 105 Pa
and a temperature of 288K. Calculate the volume of this gas at a pressure of 1.6x 105 Pa
when the temperature is 125K.
(3mks

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Approximately $2.2\; \rm cm^{3}$ (assuming that this gas is ideal, and that the initial volume of this gas is $4.0\; \rm cm^{3}$ .)

Explanation:

In this question, both the pressure on the gas and the temperature of the gas have changed. However, the ideal gas laws (Boyle's Law and Charles' Law) requires that only one of the two quantity change at a time. Therefore, consider this change in two steps:

Change pressure from $P_0 = 2.0 \times 10^{5}\; \rm Pa$ to $P_1= 1.6 \times 10^{5}\; \rm Pa$ while keeping temperature constant at $288\; \rm K$ . Find the new volume V_1

using Boyle's Law.After that, keep the pressure the same (at $1.6 \times 10^{5}\; \rm Pa$ ) and change the temperature from $T_1 = 288\; \rm K$ to $T_2 = 125\; \rm K$ . Find the final volume of the gas V_2

using Charles' Law.

Boyle's Law states that for a fixed quantity of some ideal gas, if temperature is held constant, the volume of the gas will be inversely proportional to the pressure on the gas.

Let V_0 denote the initial volume of this gas. The question states that at $P_0 = 2.0\times 10^{5}\; \rm Pa$ and $T_0 = 288\; \rm K$ , the volume of the gas is $V_0 = 4.0\; \rm cm^{3}$ .

By Boyle's Law, if temperature is held constant ( $T_1 = T_0 = 288\; \rm K$ ,) then at $P_1 = 1.6 \times 10^{5}\; \rm Pa$ :

$\begin{aligned} V_1 &= V_0 \cdot \frac{P_0}{P_1} \\ &= 4.0\; \rm cm^3 \times \frac{2.0 \times 10^{5}\; \rm Pa}{1.6 \times 10^{5}\; \rm Pa} \approx 5.0\; \rm cm^{3} \end{aligned}$ .

On the other hand, Charles' Law suggests that for a fixed quantity of some ideal gas, if the pressure of the gas is held constant, the volume of the gas will be proportional to the temperature (in degree Kelvins) of the gas.

Let V_1 denote the volume of this gas before the temperature change. At $P_0 = 2.0\times 10^{5}\; \rm Pa$ and $T_0 = 288\; \rm K$ , previous calculations show that $V_0 = 5.0\; \rm cm^{3}$ .

By Charles' Law, if the pressure of this gas is held constant ( $P_2 = P_1 =1.6 \times 10^{5}\; \rm Pa$ ,) then at the new temperature $T_2 = 125\; \rm K$ :

$\begin{aligned} V_2 &= V_1 \cdot \frac{T_1}{T_0} \\ &= 5.0\; \rm cm^3 \times \frac{125\; \rm K}{288\; \rm K} \approx 2.2\; \rm cm^{3} \end{aligned}$ .

Therefore, at $T_2 = 125\; \rm K$ and $P = 1.6 \times 10^{5}\; \rm Pa$ , the volume of this gas would be approximately $2.2\; \rm cm^{3}$ .

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

Answer from: Quest

answerscientific theory;

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9. A constant mass of hydrogen occupies a volume of 4.0 cm at a pressure of 2.0 x 105 Pa
and a...

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