For many purposes we can treat butane (C4H10) as an ideal gas at temperatures above its boiling point of - 1. °C. Suppose the pressure on a 3.0 m3 sample of butane gas at 19.0°C is reduced to one-third its initial value.
Is it possible to change the temperature of the butane at the same time such that
the volume of the gas doesn't change?

If you answered yes, calculate the new temperature of the gas. Round your
answer to the nearest °C.

The temperature of the butane can be changed when the pressure has been reduced to one-third of the initial temperature. The final temperature of the gas will be -176 .

When the pressure of butane has been reduced to one-third of the initial pressure, it has been possible to change the temperature as well. This is because the gas will follow the ideal gas equation and will be able to change pressure and temperature as they are directly proportional to each other.

The new temperature can be given by:

Initial pressure = P

Final pressure =

Initial temperature = 19

Initial temperature = 292 K

When the volume is kept constant, the relationship of the pressure and temperature can be given by:

Where P1 and T1 are initial pressure and temperature respectively. P2 and T2 are the final pressure and temperature respectively.

P 3T2 = P   292

3T2 = 292 K

T2 = 97.33 K

The final temperature of butane with the decrease of one-third of pressure has been 97.33 K.

273 K = 1

97.33 K = -176  

The final temperature of butane with the decrease of one-third of pressure has been -176  .

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A. Yes

B. –176 °C

Explanation:

A. Yes

B. Determination of the new temperature of the gas.

Let the initial pressure be P

From the question given above, the following data were obtained.

Initial pressure (P1) = P

Initial temperature (T1) = 19 °C

Final pressure (P2) = ⅓ P1 = ⅓P = P/3

Final temperature (T2) =?

Next, we shall convert 19 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T1) = 19 °C

Initial temperature (T1) = 19 °C + 273

Initial temperature (T1) = 292 K

Since the volume is constant, we can obtain the new temperature of the gas as illustrated below:

Initial pressure (P1) = P

Initial temperature (T1) = 292 K

Final pressure (P2) = P/3

Final temperature (T2) =?

P1/T1 = P2/T2

P/292 = P/3 /T2

P/292 = P/3T2

Cross multiply

P × 3T2 = 292 × P

Divide both side by P

3T2 = (292 × P)/P

3T2 = 292

Divide both side by 3

T2 = 292/3

T2 = 97.33 ≈ 97 K

Finally, we shall convert 97 K to celcius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 97

T(°C) = 97 – 273

T(°C) = –176 °C

Thus, the new temperature of the gas is –176 °C.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.

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