Question

70 cal of heat is required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30°C to 35°C. The
amount of heat required to raise the temperature of the gas through the same range at constant volume will be (assume
R=2 cal/mol-K).
(A) 30 cal
(B) 50 cal
(C) 70 cal
(D) 90 cal​

Answer 1

Answer:

• The amount of heat required is (Q) is 50 Cal

Given:

$\boxed{\begin{minipage}{12 em}\underline{\textbf{At Constant Pressure:-}}\\\\\bullet\;\sf \Delta Q_{1}=70\;cal\\\\\bullet\;\sf n=2\;mol\\\\\bullet\;\sf T_{1}=30^{\circ}C\\\\\bullet\;\sf T_{2}=35^{\circ}C\end{minipage}}\quad \boxed{\begin{minipage}{12 em}\underline{\textbf{At Constant Volume:-}}\\\\\bullet\;\sf \Delta Q_{2}=?\\\\\bullet\;\sf n=2\;mol\\\\\bullet\;\sf T_{1}=30^{\circ}C\\\\\bullet\;\sf T_{2}=35^{\circ}C\end{minipage}}$

Explanation:

$\rule{300}{1.5}$

Case - 1

Firstly lets find the Specific heat capacity at constant pressure. From the formula we know that,

$\bigstar\;\underline{\boxed{\sf \Delta Q_{1}=n\;C_{p}\;\Delta T}}$

Here,

• ΔQ₁ Denotes Heat energy.
• n Denotes moles.
• Δ T Denotes Temperature difference.

Substituting the values,

$\longrightarrow\sf 70=2\times C_{p}\times \bigg(T_{2}-T_{1}\bigg)\\\\\\\\\longrightarrow\sf 70=2\times C_{p}\times \bigg(35-30\bigg)\\\\\\\\\longrightarrow\sf 70=2\times C_{p}\times 5\\\\\\\\\longrightarrow\sf 70=10\times C_{p}\\\\\\\\\longrightarrow\sf C_{p}=\dfrac{70}{10}\\\\\\\\\longrightarrow\sf C_{p}=7\\\\\\\\\longrightarrow\sf C_{p}=7\;cal/mol\;K \quad\dots\dots\sf (1)$

Hence, we got Specific heat capacity at constant pressure.

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now lets find the Specific heat capacity at constant volume. We know,

$\longrightarrow\boxed{\sf R=C_{p}-C_{v}}\\\\\\\\\longrightarrow\sf 2=7-C_{v}\\\\\\\\\longrightarrow\sf C_{v}=7-2\\\\\\\\\longrightarrow\sf C_{v}=5\\\\\\\\\longrightarrow\sf C_{v}=5\;cal/mol\;K\quad\dots\dots\sf (2)$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Case-2

The  amount of heat required to raise the temperature of the gas through the same range at constant volume will be:-

$\bigstar\;\underline{\boxed{\sf \Delta Q_{2}=n\;C_{v}\;\Delta T}}$

Here,

• ΔQ₂ Denotes Heat energy.
• n Denotes moles.
• Δ T Denotes Temperature difference.

Substituting the values,

$\longrightarrow\sf \Delta Q_{2}=2\times5\times \bigg(T_{2}-T_{1}\bigg)\\\\\\\\\longrightarrow\sf \Delta Q_{2}=2\times 5 \times \bigg(35-30\bigg)\\\\\\\\\longrightarrow\sf \Delta Q_{2}=2\times 5 \times 5\\\\\\\\\longrightarrow\sf \Delta Q_{2}=10\times 5\\\\\\\\\longrightarrow\sf \Delta Q_{2}=50\\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf \Delta Q_{2}=50\;cal}}}}$

∴ The amount of heat required is (Q) is 50 Cal.

$\rule{300}{1.5}$