A 425-ml sample of hydrogen is collected above water at 35°c and 763 torr. find the volume of the hydrogen sample when the tempe

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A 425-ml sample of hydrogen is collected above water at 35°c and 763 torr. find the volume of the hydrogen sample when the tempe

rature falls to 23°c, assuming the barometric pressure does not change. (vapor pressures of water : at 35°c, 42.2 torr ; at 23°c, 21.1 torr)

Chemistry

2 answers:

PilotLPTM [1.2K] · Answer 1 · 2022-08-01T10:22:27+03:00

From the gas laws, Charles law states that the volume of a fixed mas of a gas is directly proportional to absolute temperature at constant pressure.
and Boyle's law states that the volume of a fixed mass of a gas is inversely proportional to pressure at constant temperature, therefore combining the two we have, PV/T, for two gases we have,
P1V1/T1=P2V2/T2
V1 = 425 ml, T1= 35 +273 =308 K, P1= 805.2 torr
V2 = ? and T2 = 23 +273 = 296K, P2 = 784.1torr
V2 = P1V1T2/P2T1
= (805.2×425×296)/784×308
= 419.486 ml
= 419.5 ml

bezimeni [28] · Answer 2 · 2022-08-01T11:33:28+03:00

Volume of hydrogen sample at $23^\circ\text{C}$ is $\boxed{419.4\text{ mL}}$ .

Further Explanation:

Ideal gas equation:

A hypothetical gas comprising of many randomly moving particles having perfectly elastic collisions between them is called ideal gas. The expression for ideal gas equation is mentioned below.

$\text{PV}=\text{nRT}$ ...... (1)

Here,

P is the pressure of gas.

V is the volume of gas.

T is the absolute temperature of gas.

n is the number of moles of gas.

R is the universal gas constant.

At temperature $\text{T}_{1}$ , volume $\text{V}_{1}$ and pressure $\text{P}_{1}$ , ideal gas equation for hydrogen gas modifies as follows:

$\text{P}_{1}\text{V}_{1}=\text{nRT}_{1}$ ...... (2)

At temperature $\text{T}_{2}$ , volume $\text{V}_{2}$ and pressure $\text{P}_{2}$ , ideal gas equation for hydrogen gas modifies as follows:

$\text{P}_{2}\text{V}_{2}=\text{nRT}_{2}$ ...... (3)

Dividing equation (3) by equation (2),

$\dfrac{\text{P}_{2}\text{V}_{2}}{\text{P}_{1}\text{V}_{1}}=\dfrac{\text{T}_{2}}{\text{T}_{1}}$ ...... (4)

Rearrange equation (4) to calculate .

$\text{V}_{2}=\dfrac{{\text{P}_{1}\text{V}_{1}\text{T}_{2}}}{\text{P}_{2}\text{T}_{1}}$ …… (5)

The value of $\text{P}_{1}$ can be calculated as follows:

$\begin{aligned}\text{P}_{1}=&763\text{ torr}+42.2\text{ torr}\\=&805.2\text{ torr}\end{aligned}$

The value of $\text{P}_{2}$ can be calculated as follows:

$\begin{aligned}\text{P}_{2}=&763\text{ torr}+21.1\text{ torr}\\=&784.1\text{ torr}\end{aligned}$

The value of $\text{T}_1$ can be calculated as follows

$\begin{aligned}\text{T}_1=&\left(35+273.15\right)\text{ K}\\=&\ 308.15\text{ K}\end{aligned}$

The value of $\text{T}_2$ can be calculated as follows

$\begin{aligned}\text{T}_2=&\left(23+273.15\right)\text{ K}\\=&\ 296.15\text{ K}\end{aligned}$

Substitute 805.2 torr for $\text{P}_{1}$ , 425 mL for $\text{V}_{1}$ , 784.1 torr for $\text{P}_{2}$ , 308.15 K for $\text{T}_{1}$ and 296.15 K for $\text{T}_{2}$ in equation (5).

$\begin{aligned}\text{V}_{2}&=\dfrac{\left({805.2\text{ torr}\right)\left(425\text{ mL}\right)\left(296.15\text{ K}\right)}}{\left(784.1\text{ torr}\right)\left(308.15\text{ K}\right)}\\&=419.4\text{ mL}\end{aligned}$