Question

In an isobaric process of a diatomic gas 140 J of
heat is added to N moles of the gas to increase
its temperature from T, to T . The amount of work
done by the gas in this process is
(1) 100 J
Q=1400
(2) 40J
(3) 60 J
(4) 120 J​

Answer 1

The amount of work  done by the gas in this process is 40 J

Explanation:

As we know that heat given in isobaric process is given as

$Q = nC_p\Delta T$

here this is diatomic gas

so we have

$C_p = \frac{7}{2}R$

now we have

$Q = n(\frac{7}{2}R)\Delta T$

now we also know that work done in isobaric process is given as

$W = nR\Delta T$

so from above equation we have

$W = \frac{2}{7}Q$

$W = \frac{2}{7}(140)$

$W = 40 J$

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Answer 2

The amount of work done by the gas in this process is 40 J.

Explanation:

For diatomic gas

$C_p=\frac{7}{2}R$

Heat absorbed=Q=140 J

The heat given in isobaric process is given by

$Q=nC_p\Delta T$

Where Q= heat

n=Number of moles

$\Delta T=$ Change in temperature

$C_p$ =Specific heat at constant pressure

Substitute the values then we get

$140=n\frac{7}{2}R\Delta T$

$\Delta T=\frac{280}{7nR}$

Work done=$nR\Delta T$

Substitute the values then we get

Work done=$nR\times \frac{280}{7nR}=40 J$

Hence, option (2) is true.

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