Question

1.13 mol sample of argon gas at a temperature of 15.0 °c is found to occupy a volume of 23.6 liters. the pressure of this gas sample is mm hg.

Answer 1

760 mmHg at 15.0 °C

Step-by-step explanation:

To solve this problem, we can use the ideal gas law, which relates the pressure (P), volume (V), number of moles (n), and temperature (T) of a gas:

• PV = nRT

where R is the universal gas constant.

We can rearrange this equation to solve for the pressure (P):

• P = nRT/V

where n, R, V, and T are given in the problem as:

• n = 1.13 mol
• R = 0.0821 L·atm/(mol·K) (the value of R in the appropriate units)
• V = 23.6 L
• T = (15.0 + 273.15) K = 288.15 K (converted to Kelvin)

Substituting these values into the equation gives:

• P = (1.13 mol)(0.0821 L·atm/(mol·K))(288.15 K)/(23.6 L)
• P = 1.00 atm

To convert this pressure to mmHg, we can use the conversion factor:

• 1 atm = 760 mmHg

Multiplying the pressure by this conversion factor gives:

• P = 1.00 atm x (760 mmHg/1 atm)
• P = 760 mmHg

Therefore, the pressure of the argon gas sample is 760 mmHg at 15.0 °C.

Answer 2

Mhm that is definitely completely correct I don’t know if I’m typing this correctly though