Answered

Exactly one mole of an ideal gas is contained in a 2.00-liter container at 1,000 K. What is the pressure exerted by this gas?

Given: R = 0.08205 L · atm/K · mol

Verified Answer :

alvincantu18

use PV =nRT

so P = nRT/V

= 1 mole(0.08205 L atm/K mol)(1000K) / 2 L

= 41 atm

Answer : The pressure of the gas is, 41.025 atm

Solution :

Using ideal gas equation :

\(PV=nRT\\\\P=\frac{nRT}{V}\)

where,

n = number of moles of gas  = 1 mole

P = pressure of the gas = ?

T = temperature of the gas = 1000 K

R = gas constant = 0.08205 L.atm/mole.K

V = volume of gas = 2.00 L

Now put all the given values in the above equation, we get the pressure of the gas.

\(P=\frac{nRT}{V}\)

\(P=\frac{1mole\times (0.08205L.atm/mole.K)\times 1000K}{2.00L}\)

\(P=41.025atm\)

Therefore, the pressure of the gas is, 41.025 atm

Related Questions