Question

If the pressure of an ideal gas contained in a closed vessel is increased by 0.5%, the increase in temperature is $2{}^{\circ }C$ . The initial temperature of the gas is:

• A. $27{}^{\circ }C$
• B. $127{}^{\circ }C$
• C. $300{}^{\circ }C$
• D. $400{}^{\circ }C$

Answer 1

Answer: (B)

Applying Gay Lussacs gas law at constant volume $P\propto T$ Hence, $\frac{P_1}{T_1}=\frac{P_2}{T_2}$ ?(i) Suppose, initial pressure $P_1=P$ Final pressure $P_2=P+\frac{0.5P}{100}$ $=1.005P$ and $T_1=T{}^{o}C$ and $T_2=(T+2){}^{o}C$ So, putting the values in Eq. (i), we get $\frac{P}{T}=\frac{1.005P}{T+2}$ $\Rightarrow$ $T+2=1.005T$ $\Rightarrow$ $1.005T=2$ $\Rightarrow$ $T=\frac{2}{1.005}=400K$ $=400-273=127{}^{o}C$