Question

20. A monatomic gas (ideal) is supplied 80 joule heat
at constant pressure. The internal energy of gas,
increases by
(1) 58 J
(2) 48 J
(3) 44 J
(4) 32J​

Answer 1

Answer:

(3) 144 J

Explanation:

Given:

Heat supplied, Q = 80 J

As given in the question the process undergoes at constant pressure, such process is called Isobaric process.

We know Heat supplied is given by

Q = nCΔT                                 -----(i)

And

Work done is given by

W = nRΔT                               -----(ii)

Now relating equation (i) and (ii) we get,

$\frac{Q}{C\Delta T} = \frac{W}{R\Delta T}$

Again we know,

$\frac{R}{C_{p}} = \frac{\gamma-1}{\gamma}$

And the value of γ for mono atomic gas, γ= 1.667

Now, after substituting all the values, we get

80 ×(1.667 - 1)/(1.667) = W

W = - 64 J ( negative sign indicates work done is by the system)

Now from Thermodynamic law, we get

Q = ΔE + W

ΔE = Q - W = 80 - (-64)

ΔE = 144 J

Therefore,

The internal energy of gas is increased by = 144 J

Answer 2

Answer:

(2) 48 J

Explanation:

We are given that a monoatomic gas(ideal) is supplied 80 joules heat at constant pressure.

We have to find the increase  value of internal energy of gas

Q=80 J

For mono atomic gas ,$C_p=\frac{5}{2}R$

$W=\frac{Q\cdot R}{C_p}=80\cdot\frac{R}{\frac{5}{2}R}}$

$W=32 J$

$\Delta E=Q-W=80-32=48 J$

Hence, the energy of gas increases by 48 J.

Answer:(2) 48 J