Question

An experimenter adds 970 J of heat to 1.75 mol of an ideal gas to heat it from ${10.0}^{\circ }C$ to ${25.0}^{\circ }C$ at constant pressure. The gas does +223 J of work during the expansion. (i) Calculate the change in internal energy of the gas. (ii) Calculate $\gamma$ for the gas.

• A. 859 J; 1.81
• B. 747 J; 2.37
• C. 650 J ; 1.29
• D. 747 J ; 1.29

Answer 1

The correct option is D

747 J ; 1.29

Listing the given information,

(i) Number of moles, n = 1.75mol,

(ii) Heat gained, Q = 970J,

(iii) Work done by the gas, W = 223J,

(iv) Rise in temperature,$\Delta T={25}^{\circ }C-{10}^{\circ }C={15}^{\circ }C$.

Therefore, from the first law, gain of internal energy -

$\Delta E_{int}=Q-W$

$\Rightarrow \Delta E_{int}=970J-223J=747J.$

We know, from the kinetic theory, $\Delta E_{int}=n{C}_V\Delta T$

$\Rightarrow C_V=\frac{\Delta E_{int}}{n\Delta T}$

= $\frac{747J}{1.75mol\times {15}^{\circ }C}$

= 28.46 J/mol.K.

Now,

$\gamma =\frac{C_P}{C_V}$

=$\frac{C_V+R}{C_V}$

= $1+\frac{R}{C_V}$

= $1+\frac{8.314}{28.46}=1.29$.