Question

A 2.37 l balloon contains 0.115 mil of xenon gas,Xe (g) at a pressure of 954 Torr

Answer 1

Answer:

The temperature in the balloon is 315 K

Explanation:

The full question

A 2.37 L balloon contains 0.115 mol of xenon gas, Xe(g), at a pressure of 954 Torr. What is the temperature, in Kelvin, of the Xe in the balloon?

Step 1: Data given

Volume of the balloon = 2.37 L

Number of moles Xenon gas = 0.115 moles

Pressure of the gas = 954 torr = 954 / 760 atm = 1.25526 atm

Step 2: Calculate the temperature of the xenon gas

p*V = n*R*T

⇒with p = the pressure of wenon gas = 1.25526 atm

⇒with V = the volume of the balloon = 2.37 L

⇒with n = the number of moles of xenon = 0.115 moles

⇒with R = the gasconstant = 0.08206 L*atm/mol*K

⇒with T = the temperature in the balloon = TO BE DETERMINED

T = (p*V) / (n*R)

T = (1.25526 * 2.37)/(0.115 * 0.08206)

T = 315 K

The temperature in the balloon is 315 K

Answer 2

Here is the full question

A 2.37 L balloon contains .115 mol of xenon gas, Xe(g), at a pressure of 954 Torr. What is the temperature, in Kelvin, of the Xe in the balloon? Express the answer using 3 significant figures.

Answer:

Temperature of Xenon = 313.77 K

Explanation:

Given that:  volume of the balloon v = 2.37 L

number of mole n  = 0.115 mol

gas constant R = 0.0821 L.atm/mol.K

temperature T = ? (in Kelvin)

pressure P = 954 torr (in atm)

since;

1 atm ----------- 760 torr

y atm ----------- 954 torr

760 y = 954

y = 954/760

y ≅ 1.25

Pressure P = 1.25 atm

Using  ideal gas equation (gas equation); we have the following expression:

$P*V = n*R*T$

Making T the subject of the formula; we have:

$T = \frac{PV}{nR}$

$T = \frac{1.25 * 2.37}{0.115 *0.0821}$

T = 313.77 K

Thus; the Temperature of Xenon = 313.77 K