5. A Dumas bulb is filled with chlorine gas at the ambient pressure and is found to contain 7.1 g of chlorine when the temperature is T (in Kelvin). The bulb is then placed in a second thermostat bath whose temperature is 300C hotter than the first one. The stopcock on the bulb is opened so that the chlorine gas pressure returns to its original value. The bulb is now found to contain 6.4 g of chlorine. (a) Calculate the value of the original temperature, T. (b) If the volume of the bulb was 2.24 liters, what was the ambient pressure?

a. The original temperature of the gas is 2743K.

b. 20atm.

Explanation:

a. As a result of the gas laws, you can know that the temperature is inversely proportional to moles of a gas when pressure and volume remains constant. The equation could be:

T₁n₁ = T₂n₂

Where T is absolute temperature and n amount of gas at 1, initial state and 2, final states.

Replacing with values of the problem:

T₁n₁ = T₂n₂

X*7.1g = (X+300)*6.4g

7.1X = 6.4X + 1920

0.7X = 1920

X = 2743K

b. Using general gas law:

PV = nRT

Where P is pressure (Our unknown)

V is volume = 2.24L

n are moles of gas (7.1g / 35.45g/mol = 0.20 moles)

R is gas constant = 0.082atmL/molK

And T is absolute temperature (2743K)

P*2.24L = 0.20mol*0.082atmL/molK*2743K

true

explanation:

who knows

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm