The pressure P of a given mass of gas varies directly as its absolute temperature T and in… — JAMB UTME

The pressure $P$ of a given mass of gas varies directly as its absolute temperature $T$ and inversely as its volume $V$ . Given that $P = 50\text{ kPa}$ when $T = 300\text{ K}$ and $V = 10\text{ m}^3$ , what is the value of $P$ when $T = 360\text{ K}$ and $V = 8\text{ m}^3$ ?

A
$48\text{ kPa}$
B
$60\text{ kPa}$
$75\text{ kPa}$ Answer
D
$90\text{ kPa}$

Answer

75\text{ kPa}

The relationship is governed by

P = \frac{kT}{V}

. Substituting

P = 50

T = 300

, and

V = 10

yields

k = \frac{5}{3}

. Using

k = \frac{5}{3}

with

T = 360

and

V = 8

gives

P = \frac{(5/3) \times 360}{8} = 75\text{ kPa}

Step-by-Step Solution

Set up the general formula for joint and inverse variation.

P = \frac{kT}{V}

, where

k

is the constant of variation.

Pressure varies directly as temperature

T

and inversely as volume

V

Substitute initial conditions to determine

k

50 = \frac{k \times 300}{10} \implies 50 = 30k \implies k = \frac{5}{3}

The initial values

P = 50\text{ kPa}

T = 300\text{ K}

, and

V = 10\text{ m}^3

allow solving for

k

Calculate the new pressure with updated temperature and volume values.

P = \frac{\frac{5}{3} \times 360}{8} = \frac{600}{8} = 75\text{ kPa}

Substitute

k = \frac{5}{3}

T = 360\text{ K}

, and

V = 8\text{ m}^3

into the variation formula.

Key Concept

Joint and inverse variation in algebraic relationships

Estimated Time:1m 30s

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