Question

in an isobaric process of the diatomic gas 140 joule of heat is added to N moles of a gas to increase its temperature from T1 to T2 amount of work done by the gas in this process is​

Answer 1

Answer:

W = 40 J

Explanation:

As we know that in isobaric process heat given is

$Q = nc_p \Delta T$

since it is given that the gas is diatomic gas

so we will have

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

$Q = n(\frac{7}{2}R)(T_2 - T_1)$

so it is given as

$140 = n(\frac{7}{2}R)(T_2 - T_1)$

now work done in isobaric process is given as

$W = n R (T_2 - T_1)$

so we have

$W = \frac{140 \times 2}{7}$

$W = 40 J$

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Topic : Isobaric Process

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Answer 2

GIVEN -Q =140 J

CP=7/2

Q= NCP(T-T°) R

ON SOLVING YOU GOT THIS ANSWER