Question

For a temperature increase of 10°C at constant volume, what is the heat absorbed by 3.0 mol of a dilute monatomic gas;

a) 30R
b) 5R
c) 45R
d)15R

Answer 1

For a temperature increase of 10°C at constant volume, the heat absorbed by 3.0 mol of a dilute monatomic gas is 45R.

Step-by-step explanation:

For a temperature increase of 10°C at constant volume, the heat absorbed by a dilute monatomic gas can be calculated using the molar heat capacity at constant volume for a monatomic ideal gas, which is given as 3/2 R per mole. Thus, for 3.0 mol of this gas, the heat absorbed (q) would be:

q = n × Cv × ΔT

Here, n is the number of moles (3.0 mol), Cv is the molar heat capacity at constant volume for a monatomic ideal gas (which is 3/2 R), and ΔT is the temperature change, which in this case is 10°C or 10 K (since a change in Celsius is equal to a change in Kelvin).

q = 3.0 mol × (3/2 R) × 10 K = 45R

So the correct answer would be (c) 45R.