A 775 ml sample of nitrogen gas is cooled from 67 $^{\circ}$ $"^@$ C to 42 $^{\circ}$ $"^@$ C. What is the new volume if the pressure remains the same?

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Answer 1 · 2017-04-03T10:19:07

The new volume is $= 718 m L$ $=718mL$

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1=V_2/T_2$

Volume, $V_{1} = 0.775 L$ $V_1=0.775L$

Temperature, $T_{1} = 67 + 273 = 340 K$ $T_1=67+273=340K$

Temperature, $T_{2} = 42 + 273 = 315 K$ $T_2=42+273=315K$

So,

$V_{2} = \frac{T_{2}}{T_{1}} \cdot V_{1}$ $V_2=T_2/T_1*V_1$

$= \frac{315}{340} \cdot 0.775 = 0.718 L$ $=315/340*0.775=0.718L$

A 775 ml sample of nitrogen gas is cooled from 67"^@∘"^@C to 42"^@∘"^@C. What is the new volume if the pressure remains the same?