Question

1.2 L of oxygen at a constant pressure of 2.00 atm.
was kept in a cylinder and provided 10.00 K cal of
heat. The volume of oxygen increases to 1.8 L.
The value of AE is​

Answer 1

Hey Dear,

◆ Answer -

∆E = 9971 cal

● Explaination -

# Given -

P = 2 atm

V1 = 1.2 L

V2 = 1.8 L

q = 10 kcal = 10000 cal

# Solution -

Work done by the gas during expansion is -

W = -P(V2-V1)

W = -2 × (1.8-1.2) × 24.2 ...(1 L.atm = 24.21 cal)

W = -29.04 cal

Change in internal energy is given by -

∆E = q + W

∆E = 10000 - 29.04

∆E = 9970.96 cal

Therefore, change in internal energy is approx 9971 cal.

Thanks dear...