Question

Two moles of an ideal monoatomic gas is heated at constant pressure of 1 atmospheric from 30 degree celsius to 90 degree celsius work done by the gases is

Answer 1

Answer:

Explanation:

internal energy = f/2 × nR∆T

for mono atomic gas f= 3

given.

n =1

and

∆T = 100-0 = 100

so

f = 3/2 ×1× 25/3 × 100

= 1250 J

Answer 2

Answer:

The work done by the gases is 997.2 Joules.

Explanation:

The work done at constant pressure is given as,

$W=P\Delta V$         (1)

Where,

W=work done

P=pressure

ΔV=change in volume

From the ideal gas equation we have,

$P\Delta V=nR\Delta T$      (2)

n=number of moles of the gas

R=universal gas constant=8.31 J.mol⁻¹.K⁻¹

ΔT=change in temperature

By putting equation (2) in equation (1) we get;

$W=nR\Delta T$       (3)

From the question we have,

Pressure=1 atmospheric

The number of moles of the gas=2

T₁=30°C

T₂=90°C

By placing all the values in equation (3) we get;

$W=2\times 8.31\times (90\textdegree C-30\textdegree C)$

$W=2\times 8.31\times 60\textdegree C$

$W=997.2 Joules$

Hence, the work done by the gases is 997.2 Joules.

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