Question

A rigid container holds 0.30g of hydrogen gas.

Answer 1

Answer:

Part A:    $\mathbf{Q =94 \ J}$ to two significant figures

Part B:    $\mathbf{Q =160 \ J}$ to two significant figures

Part C:    $\mathbf{Q =220 \ J}$ to two significant figures

Explanation:

Given that :

mass of the hydrogen = 0.30 g

the molar mass of hydrogen gas molecule = 2 g/mol

we all know that:

number of moles = mass/molar mass

number of moles = 0.30 g /2 g/mol

number of moles = 0.15 mol

For low temperature between the range of 50 K to 100 K, the specific heat at constant volume for a diatomic gas molecule = $C_v=\dfrac{3}{2}R$

For Part A:

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{3}{2})(8.314 \ J/mole.K )(100-50)K$

$Q= 0.15 \times (\dfrac{3}{2}) \times (8.314 \ J )\times (50)$

$Q=93.5325 \ J$

$\mathbf{Q =94 \ J}$ to two significant figures

Part B. For hot temperature, $C_v=\dfrac{5}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{5}{2})(8.314 \ J/mole.K )(300-250)K$

$Q= 0.15 \times (\dfrac{5}{2}) \times (8.314 \ J )\times (50)$

$Q=155.8875 \ J$

$\mathbf{Q =160 \ J}$ to two significant figures

Part C. For an extremely hot temperature, $C_v=\dfrac{7}{2}R$

$Q = mC_v\Delta T$

$Q= 0.15 \ mol (\dfrac{7}{2})(8.314 \ J/mole.K )(2300-2250)K$

$Q= 0.15 \times (\dfrac{7}{2}) \times (8.314 \ J )\times (50)$

$Q=218.2425 \ J$

$\mathbf{Q =220 \ J}$ to two significant figures