Question

The work done in heating a mole of an ideal gas at constant pressure from ${15}^{0}Cto{25}^{0}C$ is

• A. 1.987 cal
• B. 198.7 cal
• C. 9.935 cal
• D. 19.87 cal

Answer 1

The correct option is D 19.87 cal

Work done, W = P $\delta V$

$\Rightarrow W=nR\delta T$

$\Rightarrow W=1\times 8.314\times (25-15)$

$\Rightarrow W=8.314\times 10=83.14J$

$\Rightarrow W=83.14\times 0.24cal$

Therefore, W = 19.87 cal .

Hence, option D is correct.