If one mole of H₂ gas is occupies a rigid container with a capacity of 1000 litres and the temperature is raised from 27°C to 37°C, the change in pressure of the contained gas (round off to two decimal places), assuming ideal gas behaviour, is _______ Pa. (R = 8.314 J/mol-K)

The correct answer is 83.14.

Step-by-step Explanation:

We are asked to find the change in pressure ($\Delta P$) when the temperature of a gas in a rigid container is raised. Since the container is rigid, its volume remains constant throughout the process.

Let us list the given values in standard SI units:
Number of moles of gas, $n=1\text{mol}$
Volume of the container, $V=1000\text{litres}=1{\text{m}}^3$ (since $1{\text{m}}^3=1000\text{L}$)
Initial temperature, $T_1=27\text{°C}=27+273.15=300.15\text{K}$
Final temperature, $T_2=37\text{°C}=37+273.15=310.15\text{K}$
Change in temperature, $\Delta T=T_2-T_1=37-27=10\text{K}$
Ideal gas constant, $R=8.314\text{J/(mol·K)}$

According to the ideal gas equation:
$PV=nRT$

For the initial state:
$P_{1}V=nR{T}_1$

For the final state:
$P_{2}V=nR{T}_2$

Subtracting the initial state equation from the final state equation gives:
$(P_2-P_1)V=nR(T_2-T_1)$

This can be simplified in terms of the change in pressure ($\Delta P$) and change in temperature ($\Delta T$):
$\Delta P\times V=nR\Delta T$

Now, we solve for $\Delta P$ by substituting the values into the formula:
$\Delta P=\frac{nR\Delta T}{V}$

$\Delta P=\frac{1\times 8.314\times 10}{1}$

$\Delta P=83.14\text{Pa}$

Thus, the change in pressure of the contained gas is 83.14 Pa.