Question

A 75.0-liter canister contains 15.82 moles of argon at a pressure of 546.8 kilopascals. What is the temperature of the canister?

Answer 1

Answer:

The absolute temperature, T of the canister is 311.44 K.

Explanation:

Given data:

Volume of Argon in the canister, V = 75.0 litre

No. of moles of argon, n = 15.82

Pressure, P = 546.8 kPa=546.8/101.325atm=5.39 atm   ..... [ ∵ 1 atm = 101.325kPa]

To find: Temperature, T of the canister

Using the Ideal Gas Law, we get

PV = nRT ….. (i)

Where, P = pressure, V= volume, n= no.of moles, T= absolute temperature and R = ideal gas constant = 0.082057 L atm mol^-1K^-1

By substituting the given values in (i), we get

5.39 * 75 = 15.82 * 0.082057 * T

Or, T = 404.25 / 1.298 = 311.44 K