Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 1151

If a quantity of heat \(1163.4 \mathrm{~J}\) is supplied to one mole of nitrogen gas, at room temperature at constant pressure, then the rise in temperature is \(\mathrm{R}=8.31\\{\mathrm{~J} /(\mathrm{m} \cdot 1 . \mathrm{k})\\}\) (A) \(28 \mathrm{~K}\) (B) \(65 \mathrm{~K}\) (C) \(54 \mathrm{~K}\) (D) \(40 \mathrm{~K}\)

Short Answer

Expert verified
The rise in temperature when a quantity of heat (1163.4 J) is supplied to one mole of nitrogen gas at room temperature at constant pressure is approximately 40 K. Therefore, the correct option is (D) \(40\,\mathrm{K}\).

Step by step solution

01

Write down the formula for heat capacity at constant pressure

We will use the formula for heat capacity at constant pressure (Cp) for an ideal gas: \[ q = nC_p∆T \] where q = heat supplied (1163.4 J), n = number of moles (1 mole), Cp = heat capacity at constant pressure, ∆T = rise in temperature.
02

Determine the heat capacity at constant pressure for nitrogen gas

For diatomic molecules like nitrogen gas, the molar heat capacity at constant pressure (Cp) can be calculated by adding the gas constant R to the molar heat capacity at constant volume (Cv), which would result in the equation: \[ C_p = C_v + R \] For a diatomic molecule, the molar heat capacity at constant volume (Cv) is given by the equation: \[ C_v = \dfrac{5}{2} R \] Substitute the value of Cv in Cp formula: \[ C_p = \dfrac{5}{2} R + R \]
03

Calculate Cp using the given value of R

Now, substitute the given value of R in the Cp formula: \[ C_p = \dfrac{5}{2}(8.31) + (8.31) \] \[ C_p = 20.775 + 8.31 = 29.085 \, \mathrm{J/(mol.K)} \]
04

Solve for the rise in temperature (∆T)

Now, we have all the necessary values to find the rise in temperature (∆T). The formula that we will use is: \[ ∆T = \dfrac{q}{nC_p} \] Substitute the given values: \[ ∆T = \dfrac{1163.4 \,\mathrm{J}}{1\,\mathrm{mol} \times 29.085\, \mathrm{J/(mol.K)}} \]
05

Calculate the final answer

Calculate the final value of the rise in temperature: \[ ∆T = \dfrac{1163.4}{29.085} = 39.96 \approx 40\,\mathrm{K} \] The rise in temperature when a quantity of heat (1163.4 J) is supplied to one mole of nitrogen gas at room temperature at constant pressure is approximately 40 K. Therefore, the correct option is (D) \(40\,\mathrm{K}\).

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

The first law of thermodynamics is concerned with the conservation of (A) momentum (B) energy (C) mass (D) temperature

The temperature of sink of Carnot engine is \(27^{\circ} \mathrm{C}\). Efficiency of engine is \(25 \%\) Then find the temperature of source. (A) \(227^{\circ} \mathrm{C}\) (B) \(327^{\circ} \mathrm{C}\) (C) \(27^{\circ} \mathrm{C}\) (D) \(127^{\circ} \mathrm{C}\)

One mole of an ideal gas $\left(\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right)=\gamma$ at absolute temperature \(\mathrm{T}_{1}\) is adiabatically compressed from an initial pressure \(\mathrm{P}_{1}\) to a final pressure \(\mathrm{P}_{2}\) The resulting temperature \(\mathrm{T}_{2}\) of the gas is given by. (A) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{\gamma /(\gamma-1)\\}}$ (B) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\\{(\gamma-1) / \gamma\\}}$ (C) $\mathrm{T}_{2}=\mathrm{T}_{1}\left\\{\mathrm{p}_{2} / \mathrm{p}_{1}\right\\}^{\gamma}$ (D) $\mathrm{T}_{2}=\mathrm{T}_{1}\left(\mathrm{p}_{2} / \mathrm{p}_{1}\right)^{\gamma-1}$

A gas mixture consists of 2 mole of oxygen and 4 mole of argon at temperature \(\mathrm{T}\). Neglecting all vibrational modes, the total internal energy of the system is (A) \(11 \mathrm{RT}\) (B) \(9 \mathrm{RT}\) (C) \(15 \mathrm{RT}\) (D) \(4 \mathrm{RT}\)

For hydrogen gas \(\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{a}\) and for oxygen gas \(\mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=\mathrm{b}\), The relation between \(\mathrm{a}\) and \(\mathrm{b}\) is given by (A) \(a=4 b\) (B) \(a=b\) (C) \(\mathrm{a}=16 \mathrm{~b}\) (D) \(a=8 b\)
See all solutions

Recommended explanations on English Textbooks

Phonology

Read Explanation

Argumentative Essay

Read Explanation

Morphology

Read Explanation

English Grammar

Read Explanation

Semiotics

Read Explanation

Text Comparison

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