Question

4 moles of a perfect gas heated to increase its temperature by 2 degree Celsius absorbs heat of 40 calorie at constant volume if the same gas is heated at constant pressure the amount of heat supplied is​

Answer 1

heat at constant volume , U = 40 calorie

or, nCv∆T = 40 calorie

or, 4mol × Cv × 2°C = 40 calorie

or, Cv = 5 Calorie/mol.°C

we know, Cv = R/(1 - Y)

R is universal gas constant. i.e., R = 2 cal/mol.°C

so, 5 = 2/(1 - Y)

or, 5(1 - Y) = 2

or, 5 - 5Y = 2

or, 3 = 5Y

or, Y = 5/3

now, Cp = YR/(1 - Y) = 5/3 × 5 = 25/3 cal/mol.°C

so, heat at constant pressure = nCp∆T

= 4 mol × 25/3 cal/mol.°C × 2 °C

= 200/3 calorie

= 66.67 calorie

Answer 2

Answer: Given that,

n=2 moles, dT=4 degree Celsius, dQ=32cal, R=2cal/mol K

Explanation:

At Constant Volume,

dQ=nCvdT=dU=32

At Constant Pressure,

dQ=dU+dW

dQ=dU+nRdT

dQ=32+(4×2×2)

dQ=32+16

dQ=48

The amount of heat supplied is 48 cal

Hope it helps you