Question

when 1210 j of heat are added to one - mole of an ideal monatomic gas, its temperature increases from 272 k to 275 k. find the amount of work done by the gas during this process.

a. 5617 j

b. 4552 j

c. 1160 j

d. 2423 j

e. 7685 j

Explanation:

Step1: Use the internal - energy formula for an ideal monatomic gas

The internal - energy change of an ideal monatomic gas is given by $\Delta U=\frac{3}{2}nR\Delta T$. Here, $n = 1$ mole, $\Delta T=T_2 - T_1=275 - 272=3\ K$.

Step2: Calculate $\Delta U$

$\Delta U=\frac{3}{2}\times1\times R\times3$. Since $R = 8.31\ J/(mol\cdot K)$, $\Delta U=\frac{3}{2}\times8.31\times3=37.395\ J$.

Step3: Apply the first - law of thermodynamics

The first - law of thermodynamics is $\Delta U = Q - W$, where $Q$ is the heat added to the system and $W$ is the work done by the system. We know that $Q = 1210\ J$ and $\Delta U$ is calculated above. Rearranging for $W$, we get $W=Q-\Delta U$.
$W = 1210-\frac{3}{2}\times8.31\times3=1210 - 37.395 = 1160\ J$ (approx).

Answer:

C. 1160 J