Question

A gas expands from a volume of 3.0 $d{m}^3$ to 5 $d{m}^3$ against a constant pressure of 3.0 atm. The work done during expansion is used to heat 10.0 mole of water of temperature 290.0 K. Calculate the final temperature of water. (Specific heat of water = 4.184 $J{K}^{-1}{g}^{-1}$

• A. 289.193 K
• B. 290.807 K
• C. 304.526 K
• D. 275.470 K

Answer 1

The correct option is B 290.807 K

Work done = - P$\times$dV
=-$3.0\times (5.0-3.0)$
= -$6.0L-atm$
= -$6.0\times 101.3J$
= -607.8 J
Magnitude of work done = 607.8 J
Consider $\Delta T$ be the change in temperature during the process
Heat absorbed = $m\times s\times \Delta T$
$=10.0\times 18\times 4.184\times \Delta TJ$
According to the given question
Work done = Heat absorbed
$\Rightarrow$ 607.8 J = $m\times s\times \Delta T$
= $\Delta T=\frac{607.8}{m\times s}$
= $\Delta T=\frac{607.8}{10.0\times 18\times 4.184}$
$\Rightarrow \Delta T=0.807K$
Hence final temperature = 290 + 0.807
= 290.807 K