Question

The temperature of 4.29 mol of an ideal diatomic gas is increased by 27.1 ˚C without the pressure of the gas changing. The molecules in the gas rotate but do not oscillate. (a) How much energy is transferred to the gas as heat? (b) What is the change in the internal energy of the gas? (c) How much work is done by the gas? (d) By how much does the rotational kinetic energy of the gas increase?

Answer 1

Answer:

energy is transferred to the gas is 37.46 ×10³ J

change in the internal energy is 26.75 ×10³ J

work done = 10.70 ×10³ J

the rotational kinetic energy is 16.06 ×10³ J

Explanation:

given data

mole = 4.29 mol

temperature = 27.1˚C = 300.1 k

to find out

How much energy is transferred to the gas , change in the internal energy, and work is done by the gas and how much does the rotational kinetic energy

solution

we know here energy formula for constant pressure that is

Q = n×Cp×Δt   .............1

and for ideal gas and diatomic process

Q = $\frac{7}{2}$×n×R×Δt

here Q is energy and Δt is change in temperature and n is no of mole and R is 8.314 J/mol-K

so

Q = $\frac{7}{2}$×4.29×8.314×300.1

Q = 37.46 ×10³ J

and

change in internal energy is

ΔE = Q -W

ΔE =  $\frac{7}{2}$×n×R×Δt  - nRΔt

ΔE =  $\frac{5}{2}$×n×R×Δt

ΔE =  $\frac{5}{2}$×4.29×8.314×300.1

ΔE =  26.75 ×10³ J

and

work done is express as

work done = nRΔt

work done = 4.29×8.314×300.1

work done = 10.70 ×10³ J

and

change in total transnational energy is

ΔE = ΔK

for diatomic and motion

ΔK =  $\frac{3}{2}$×n×R×Δt

ΔK =  $\frac{3}{2}$×4.29×8.314×300.1

ΔK =  16.06 ×10³ J