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Chapter 19: Q14P (page 578)

Question: In the temperature range 310 K to 330 K, The pressure P of a certain non ideal gas is related to volume V and temperature T by $p = (24.9 \frac{J}{K}) \frac{T}{V} - (0.00662 \frac{J}{K^{2}}) \frac{T^{2}}{V}$
How much work is done by the gas if its temperature is raised from 315 Kto 325 Kwhile the pressure is held constant?

Short Answer

Expert verified

Answer

Work done by the gas is 207 J.

Step by step solution

01

Step 1: Given

$T_{1} = 315 K T_{2} = 325 K P = (24.9) \frac{T}{V} - (0.00662) \frac{T^{2}}{V}$

02

Determine the concept

Formula is as follow:

$W = P_{螖} V$

Here, Pis pressure, V is volume and W is work done.

03

Determine the work done by the gas

Write the expression for pressure and volume as:

$P = (24.9) \frac{T}{V} - (0.00662) \frac{T^{2}}{V} 鈭� V = (24.9) \frac{T}{P} - (0.00662) \frac{T^{2}}{P}$

Solve for the work done as:

$W = P 鈻� V W = P ((24.9) \frac{T}{P} - (0.00662) \frac{T^{2}}{P}) W = P ((24.9) \frac{鈻� T}{P} - (0.00662) \frac{鈻� (T^{2})}{P}) W = P ((24.9) \frac{(T_{2} - T_{1})}{P} - (0.00662) \frac{T_{2}^{2} - T_{1}^{2}}{P})$

Substitute the values and solve as:

$W = P ((24.9) \frac{(325 - 315)}{P} - (0.00662) \frac{(325^{2} - 315^{2})}{P}) W = ((24.9) (325 - 315) - (0.00662) (325^{2} - 315^{2})) W = 206.632 鈮� 207 J$

Hence, thework done by the gas is 207 J.

Therefore, the work done by the gas can be found by using the formula for work done in isobaric process.

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Most popular questions from this chapter

For four situations for an ideal gas, the table gives the energy transferred to or from the gas as heat Qand either the work W done by the gas or the work $W_{o n}$ done on the gas, all in joules. Rank the four situations in terms of the temperature change of the gas, most positive first.

Figure shows two paths that may be taken by a gas from an initial point i to a final point f. Path 1 consists of an isothermal expansion (work is $50 鈥塉$ in magnitude), an adiabatic expansion (work is $40 鈥塉$ in magnitude), an isothermal compression (work is 30Jin magnitude) and then an adiabatic compression (work is 25Jin magnitude). What is the change in the internal energy of the gas if the gas goes from point i to point f along path 2?

The dot in Fig 19-18 a represents the initial state of a gas, and the vertical line through the dot divides the p-V diagram into regions 1 and 2. For the following processes, determine whether the work W done by the gas is positive, negative, or zero: (a) the gas moves up along the vertical line, (b) it moves down along the vertical line, (c) it moves to anywhere in region 1, and (d) it moves to anywhere in region 2.

Question: The lowest possible temperature in outer space is 2.7 K. What is the rms speed of hydrogen molecules at this temperature? (The molar mass is $2.02 脳 10^{- 3} k g / m o l$ ).

The temperature of $2.00$ of an ideal monatomic gas is raised $15.0 鈥� K$ in an adiabatic process. What are

(a) the work W done by the gas,

(b) the energy transferred as heat Q,

(c) the change ${螖贰}_{int}$ in internal energy of the gas, and

(d) the change $螖碍$ in the average kinetic energy per atom?

See all solutions

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