Question

5 g of a gas is contained in a rigid container and is heated from ${15}^{0}C$ to ${25}^{0}C$. Specific heat capacity of the gas at constant volume is 0.172 cal $g{-1}^C-1$ and the mechanical equivalent of heat is 4.2 J $ca{l}^{-1}$ Calculate the change in the internal energy of the gas.

Answer 1

m = 5 g,

$\Delta T=25-{15}^{0}C={10}^C$

$C_v=0.172cal/g{-}^{\circ }C,$

J = 4.2 J/cal

We know,

dQ = dU+dW

Now, V= 0 (for a rigid body)

So, dW = 0

So, dQ = dU

Q = msdT = $5\times 0.172\times 10$

= 8.6 cal = 8.6 $\times 4.2$ J

= 36.12 J.