Answer
308 K
Work Step by Step
According to the ideal gas law, $PV=nRT$ since n is constant, $n_{1}R=n_{2}R.$ So, $P_{1}V_{1}/T_{1}=P_{2}V_{2}/T_{2}$.
Let's take,
The cross-sectional area of the beaker = A
The height of the region occupied by the gas = h
We can write,
$$\frac{P_{1}Ah_{1}}{T_{1}}=\frac{P_{2}Ah_{2}}{T_{2}}-(1)$$
Let's find the $P_{1}, P_{2}$ as follows.
$P_{1}=P_{atm}+\rho gH$ ; Let's plug known values into this equation.
$P_{1}=1.01\times10^{5}Pa+(1.36\times10^{4}kg/m^{3})(9.8\space m/s^{2})(0.76\space m)$
$P_{1}=2.02\times10^{5}Pa$
$P_{2}=P_{atm}+\rho gH$ ; Let's plug known values into this equation.
$P_{2}=1.01\times10^{5}Pa+(1.36\times10^{4}kg/m^{3})(9.8\space m/s^{2})(0.38\space m)$
$P_{2}=1.52\times10^{5}Pa$
From (1)=>
$T_{2}=(\frac{P_{2}h_{2}}{P_{1}h_{1}})T_{1}$ ; Let's plug known values into this equation.
$T_{2}=\frac{(1.52\times10^{5}Pa)(1.14\space m)}{(2.02\times10^{5}Pa)(0.76\space m)}(273\space K)=308\space K$
