#### Calculate the % increase in temperature of gas, when it is heated at constant pressure to occupy 20% increase in volume.Option: 1 30Option: 2 15Option: 3 20Option: 4 25

We have:

$\\\mathrm{V_{2}\, =\,V\, +\, \frac{20V}{100}}\: =\: \frac{120V}{100}$

$T=T_2$

From Charles’ law, we know:

$\\\mathrm{\frac{V_{1}}{T_{1}} =\frac{V_{2}}{T_{2}}}\\\\\mathrm{\therefore \: T_{2}\: =\: \frac{120V\, x\, T_{1}}{100\, x\, V}\: =\: 1.2T_{1}}$

Thus, increase in temperature $=1.2T_1-T_1=0.2T_1$

Therefore, % increase in temperature $=(0.2T_1/T_1)\times 100=20\%$

Alternate Solution:

$\begin{array}{l} \text { Vol }^m \text { Initial }=V \\ \text { Increase in\ vol }^{m}=V+V \times \frac{20}{100}=1.2V \end{array}$

$\begin{array}{l}\because \text { At constant pressure } \\ \qquad \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} \end{array}$

$\mathrm{\Rightarrow \frac{V}{T_{1}}=\frac{1.2V}{T_{2}}}$

$\\\mathrm{\Rightarrow \frac{T_2}{T_{1}}=1.2} \\\\\mathrm{\Rightarrow \frac{T_2}{T_{1}}-1=1.2-1 =0.2}$

$\\\\\mathrm{\Rightarrow \frac{T_2-T_{1}}{T_1}\times 100=20\%}$