Question

One mole of an ideal gas (Cv = 20J K-'mol-') initially at STP is heated at constant
volume to twice the initial temperature. For the process, W and q will be :
1) W = 0;9 = 5.46kJ
2) W = 0;q = 0
3) W = -5.46 kJ;9 = 5.46 kJ
4) W = 5.46 kJ;q = 5.46kJ​

Answer 1

Given info : One mole of an ideal gas (Cv = 20J K-¹mol-¹) initially at STP is heated at constant

volume to twice the initial temperature.

To find : For the process, W and q will be ...

solution : gas is headed at constant volume.

so, volume remains same

it means, work done , W = 0 [ as W = P∆V]

from first law of thermodynamics,

q = U + W

⇒q = U

⇒q = nCv∆T

= 1 mol × 20 J/mol/K × (2T - T) ,

where T is standard temperature. i.e., T = 273K

= 20 × 273 J

= 5460 J = 5.46 kJ

Therefore the correct option is (1) W = 0, q = 5.46 kJ.