Question

2 moles of an ideal gas at temperature 27°C is
heated isothermally from volume V to 4V. If R = 2
cal/mol K, then the heat input in this process is
approximately​

Answer 1

isothermal process ; ∆T = 0

so, change in internal energy , ∆U = 0

from first law of thermodynamics,

∆Q = ∆U + W

⇒∆Q = 0 + W

⇒∆Q = W

hence, heat input in this process is equal to workdone by gas.

Work done , W = P∫dV

we know, PV = nRT ⇒P = nRT/V

so, W =nRT ∫dV/V

= nRTln[V2/V1]

so, ∆Q = nRTln[V2/V1]

here, n = 2, R = 2Cal/mol.K and T = 27° = 27 + 273 = 300K , V2 = 4V, V1 = V

so, ∆Q = 2 × 2 × 300 × ln[4V/V]

= 1200ln(4)

= 2400ln(2)

= 2400 × 0.693

= 1663.2 Cal

Answer 2

Answer:

See the attachment

Good luck

Explanation: