What would be the molar volume of one mole of $H_{2}$ gas if the STP has been changed to $27^{0}$ C and $2$ atm pressure?

24.62

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22.4

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12.315

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6.15

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Solution

The correct option is C $12.315$ L
$\begin{matrix} Given T = 27^{\circ} C = 27 + 273 = 300 K P = 2 a t m V_{m} = ? at STP n H_{2} = 1, R = 0.0821 \frac{a t m L}{K} \end{matrix}$

$\begin{matrix} Ideal gas equation, \begin{matrix} P V = & n R T V_{m} = \frac{M R T}{P} & = \frac{1 \times 0.0821 \times 300}{2} = \frac{24.63}{2} V m & = 12.315 L \end{matrix} \end{matrix}$

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What would be the molar volume of one mole of H2 gas if the STP has been changed to 270 C and 2 atm pressure?

The correct option is C 12.315 L Given T=27∘C=27+273=300 KP=2 atmVm=? at STP nH2=1,R=0.0821atmLK Ideal gas equation, PV=nRTVm=MRTP=1×0.0821×3002=24.632Vm=12.315L

What would be the molar volume of one mole of $H_{2}$ gas if the STP has been changed to $27^{0}$ C and $2$ atm pressure?