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Question: In an isobaric process, the work done by a di-atomic gas is 10J, the heat given to the gas will be? ...

In an isobaric process, the work done by a di-atomic gas is 10J, the heat given to the gas will be?
A. 35J
B. 30J
C. 45J
D. 60J

Explanation

Solution

The pressure remains constant for an isobaric process. We will use the formula of work done to find the difference in temperature. Then we will use that value to find the heat needed to raise the temperature of the gas by that amount under isobaric conditions.
Formula used:
W=P.ΔVW=P.\Delta V
Q=nCpΔTQ=n{{C}_{p}}\Delta T
Cp=Cv+R{{C}_{p}}={{C}_{v}}+R
Cv=fR2{{C}_{v}}=\dfrac{fR}{2}

Complete answer:
Using the ideal gas equation, we can write PV=nRTPV=nRT. Here the pressure remains constant throughout so the formula for the work done will be given by W=P.ΔVW=P.\Delta V. Now as the number of moles remain constant this can be written to be equal to W=nRΔTW=nR\Delta T. Here, n is the number of moles of the given gas in the given amount. R is the universal gas constant and ΔT\Delta T is the difference in temperature. The work done is given to be equal to 10 Joules. Using this we get the value of ΔT\Delta T to be nRΔT=10ΔT=10nRnR\Delta T=10\Rightarrow \Delta T=\dfrac{10}{nR}. Now the given gas is diatomic in nature. The degree of freedom for a diatomic gas is 5. The value of specific heat constant for diatomic gas at constant volume is given as Cv=fR2{{C}_{v}}=\dfrac{fR}{2}. Here f is the degrees of freedom of the molecules of the gas and R is the universal gas constant. And the specific heat at constant pressure is given as Cp=Cv+R{{C}_{p}}={{C}_{v}}+R. So Cp=fR2+R=5R2+R=7R2{{C}_{p}}=\dfrac{fR}{2}+R=\dfrac{5R}{2}+R=\dfrac{7R}{2}. Using this value for the specific heat we get the value of heat given to the gas as Q=nCpΔT=n×7R2×10nR=7×102=35JQ=n{{C}_{p}}\Delta T=n\times \dfrac{7R}{2}\times \dfrac{10}{nR}=\dfrac{7\times 10}{2}=35J.

Hence, the correct option is A, i.e. 35 J.

Note:
Students must take care that the formula for work done may not always be the same as the pressure may not be isobaric and an integral formula may have to be used. Also, the degrees of freedom must be written according to the nature of the gas given.