Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 1112

70 calorie of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from \(30^{\circ} \mathrm{C}\) to $35^{\circ} \mathrm{C}$ The amount of heat required to raise the temperature of the same gas through the same range at constant volume is $\ldots \ldots \ldots \ldots \ldots .$ calorie. (A) 50 (B) 30 (C) 70 (D) 90

Short Answer

Expert verified
The heat required to raise the temperature of the same gas through the same range at constant volume is approximately 50 calories. The correct answer is (A) 50.

Step by step solution

01

Identify the given information

We are given that 70 calories of heat are required to raise the temperature of 2 moles of an ideal gas from \(30^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C}\) at constant pressure. Also, we need to find the heat required to raise the temperature of the gas through the same range at constant volume.
02

Use the heat formula at constant pressure

We can use the formula for heat at constant pressure: \[Q_p = nC_p \Delta T\] where \(Q_p\) is the heat at constant pressure, \(n\) is the number of moles, \(C_p\) is the specific heat capacity at constant pressure, and \(\Delta T\) is the change in temperature.
03

Calculate the specific heat capacity at constant pressure

We can plug in the given values and solve for \(C_p\): \[70 \text{ cal} = 2 \text{ moles} \cdot C_p \cdot (35^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C})\] First, calculate the change in temperature: \(\Delta T = 35^{\circ} \mathrm{C} - 30^{\circ} \mathrm{C} = 5^{\circ} \mathrm{C}\) Now, plug it into the equation to find \(C_p\): \[C_p = \frac{70 \text{ cal}}{2 \text{ moles} \cdot 5^{\circ} \mathrm{C}} = 7 \frac{\text{cal}}{\text{mole}\cdot ^{\circ} \mathrm{C}}\]
04

Find the specific heat capacity at constant volume

For an ideal gas, the relationship between specific heat capacities at constant pressure and constant volume is: \[C_p - C_v = R\] where \(R\) is the gas constant. For this problem, we can use the value of \(R = 1.987 \frac{\text{cal}}{\text{mole}\cdot ^{\circ} \mathrm{C}}\). Now, we can find \(C_v\): \[C_v = C_p - R = 7 \frac{\text{cal}}{\text{mole}\cdot ^{\circ} \mathrm{C}} - 1.987 \frac{\text{cal}}{\text{mole}\cdot ^{\circ} \mathrm{C}} = 5.013 \frac{\text{cal}}{\text{mole}\cdot ^{\circ} \mathrm{C}}\]
05

Calculate the heat at constant volume

Now that we have \(C_v\), we can calculate the heat at constant volume using the formula: \[Q_v = nC_v \Delta T\] Plug in the values: \[Q_v = 2 \text{ moles} \cdot 5.013 \frac{\text{cal}}{\text{mole}\cdot ^{\circ} \mathrm{C}} \cdot 5^{\circ} \mathrm{C}\] Calculate the heat at constant volume: \[Q_v = 50.13 \text{ cal}\] Since the given options are all integers, we can round the answer to the nearest integer: \[Q_v \approx 50 \text{ cal}\] So, the heat required to raise the temperature of the same gas through the same range at constant volume is approximately 50 calories. The correct answer is (A) 50.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\mathrm{Cp}\) and Cv denote the specific heat of oxygen per unit mass at constant Pressure and volume respectively, then (A) \(\mathrm{cp}-\mathrm{cv}=(\mathrm{R} / 16)\) (B) \(\mathrm{Cp}-\mathrm{Cv}=\mathrm{R}\) (C) \(\mathrm{Cp}-\mathrm{Cv}=32 \mathrm{R}\) (D) \(\mathrm{Cp}-\mathrm{Cv}=(\mathrm{R} / 32)\)

An engine is supposed to operate between two reservoirs at temperature \(727^{\circ} \mathrm{C}\) and \(227^{\circ} \mathrm{C}\). The maximum possible efficiency of such an engine is (A) \((3 / 4)\) (B) \((1 / 4)\) (C) \((1 / 2)\) (D) 1

One mole of a monoatomic gas \([\gamma=(5 / 3)]\) is mixed with one mole of A diatomic gas \([\gamma=(7 / 5)]\) what will be value of \(\gamma\) for mixture ? (A) \(1.454\) (B) \(1.4\) (C) \(1.54\) (D) \(1.5\)

In an isochoric Process \(\mathrm{T}_{1}=27^{\circ} \mathrm{C}\) and \(\mathrm{T}_{2}=127^{\circ} \mathrm{C}\) then $\left(\mathrm{P}_{1} / \mathrm{P}_{2}\right)$ will be equal to (A) \((9 / 59)\) (B) \((2 / 3)\) (C) \((4 / 3)\) (D) \((3 / 4)\)

An ideal gas at \(27 \mathrm{C}\) is Compressed adiabatically, to \(\\{8 / 27\\}\) of its original Volume. If \(\mathrm{v}=(5 / 3)\), then the rise in temperature is (A) \(225 \mathrm{k}\) (B) \(450 \mathrm{~K}\) (C) \(375 \mathrm{~K}\) (D) \(405 \mathrm{~K}\)
See all solutions

Recommended explanations on English Textbooks

Graphology

Read Explanation

Research and Composition

Read Explanation

Single Paragraph Essay

Read Explanation

Email

Read Explanation

Creative Story

Read Explanation

Semiotics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free