Solve.
-The volume $\mathrm { V }$ of a given mass of gas varies directly as the temperature $\mathrm { T }$ and inversely as the pressure $\mathrm { P }$. A measuring device is calibrated to give $\mathrm { V } = 319.2 \mathrm { in } ^ { 3 }$ when $\mathrm { T } = 570 ^ { \circ }$ and $\mathrm { P } = 25 \mathrm { lb } / \mathrm { in } ^ { 2 }$. What is the volume on this device when the temperature is $430 ^ { \circ }$ and the pressure is $20 \mathrm { lb } / \mathrm { in } ^ { 2 }$ ?
A) $V = 311$ in $^ { 3 }$
B) $\mathrm { V } = 21.5 \mathrm { in } ^ { 3 }$
C) $V = 301$ in $^ { 3 }$
D) $V = 291$ in $^ { 3 }$

Question

Quizplus · Accepted Answer

The volume of a gas varies directly as the temperature and inversely as the pressure, which can be represented by the formula $V = k \frac{T}{P}$, where $k$ is a constant. To find $k$, use the given conditions $V = 319.2$ in$^3$, $T = 570$°, and $P = 25$ lb/in$^2$. Solving for $k$, we get $k = \frac{V \cdot P}{T} = \frac{319.2 \cdot 25}{570}$. Then, to find the new volume $V$ when $T = 430$° and $P = 20$ lb/in$^2$, substitute these values and the calculated $k$ into the formula. The calculation shows that the volume is closest to $301$ in$^3$, making option C correct.

Solve -The Volume V\mathrm { V }V Of a Given Mass of Gas Varies Directly as the Mass

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-The Volume $\mathrm { V }$ Of a Given Mass of Gas Varies Directly as the Mass